ch04

Chapter 4

The Energy of Particles

4.1 Kinetic Energy

75

4.1.1GOAL: Find the work performed by the lift.GIVEN: m = 1100 kg, y = 0.9 m/s2, y

2− y

1= 2.0 m

DRAW

FORMULATE EQUATIONS: The work performed from state 1 to state 2 is given by

W =∫ y

2

y1

Ndy (1)

The balance of forces on the car is

N −mg = my ⇒ N = m(g + y) (2)

SOLVE:(2) → (1) ⇒ W = m(g + y)(y

2− y

1)

W = (1100 kg)(9.81 m/s2 + 0.9 m/s2)(2.0 m) = 23, 562 N·m

W = 23, 562 N·m

76

4.1.2GOAL: Find the speed of the mass when it is 0.6 m from the wallGIVEN: m = 0.5 kg, k = 40N/m, unstretched length=L

0= 0.3 m, x

1= 2m, x

2= 0.6 m

DRAW:

ASSUME: The surface is frictionless and the spring is linear.FORMULATE EQUATIONS:We’ll apply work/energy:

KE∣∣∣2

= KE∣∣∣1+ W

∣∣∣1−2

KE∣∣∣1

= 0 ⇒ KE∣∣∣2

= W∣∣∣1−2

SOLVE:W∣∣∣1−2

=∫ s

2

s1

Ftdt =∫ x

2

x1

−(k(x− L0))dx

W∣∣∣1−2

= −k

∫ x2

x1

(x− L0)dx = −k[

x2

2− L

0x]∣∣∣x2

x1

W∣∣∣1−2

= −(40 N/m)[((0.6 m)2

2−(0.3 m)(0.6 m) )−(

(2 m)2

2−(0.3 m)(2m) )] = 56 J

KE∣∣∣2

= W∣∣∣1−2

= 56 J =12mv2 ⇒ v = 14.97 m/s

⇀v = −14.97⇀ı m/s

77

4.1.3GOAL: Find speed of arrow after it has moved 1.6 feetGIVEN: arrow’s weight and force profileDRAW:

FORMULATE EQUATIONS:

KE∣∣∣2

= KE∣∣∣1+ W

1−2

SOLVE:

m =(20 oz)

(16 oz/lb)(32.2 ft/s2)= 3.88× 10−3 slug

KE∣∣∣1

= 0

W1−2

= [1.6∫

0

40e−3.2xdx]ft·lb

=

− 403.2

e−3.2x

∣∣∣∣∣1.6

0

ft·lb = 12.4 ft·lb

12

mx2 = W1−2

12

(3.88× 10−3 slug)x2 = 12.4 ft·lb

x = 80 ft/s

78

4.1.4GOAL: Find the speed of a mass after being acted on by a given force.GIVEN: Initial speed, mass, force profileDRAW:


KE∣∣∣2

= KE∣∣∣1+ W

1−2

SOLVE:

KE∣∣∣1

=12

mv2 =12

(50 kg)(3 m/s)2 = 225N·m

W1−2

= −2∫

0

[−100− 50e−1.1x

(2 + x)]dx

Evaluating with the MATLAB M-file quad yields W1−2

= −113 N·m

12

(50 kg)x2 = 225N·m− 113 N·m

x = 2.12 m/s

79

4.1.5GOAL: Plot speed and acceleration of a cyclist going downhillGIVEN: Mass, slope, and drag forcesDRAW:

⇀ı ⇀⇀b1 cos θ sin θ⇀b2 − sin θ cos θ


Force balance: −mg⇀ + F⇀b1 + N

⇀b2 = −ms

⇀b1

−mg(sin θ⇀b1 + cos θ

⇀b2) + (5 N + as2)

⇀b1 + N

⇀b2 = −ms

⇀b1

⇀b1 : ms = mg sin θ − 5 N− as2 (1)

⇀b2 : −mg cos θ + N = 0 (2)

Work/displacement:∫ s2

s1Fds = W1−2 (3)

SOLVE:

(1) ⇒ s = g sin θ − 5 Nm

− 0.04s2

m(4)

Using (4) in MATLAB (θ = 6◦, a = 0.04 N · s2 /m2, m = 50 kg) produces the required s, s data.Using this data along with (4) then lets us solve for s and produce the following two plots.

80

To calculate the work done by drag forces we employ

Wdrag = −∫ 50m

0

(5 N + as2

)ds

Using the speed/displacement data already obtained from the numerical simulation lets us calculatethe work done and produces a final result of

Wdrag

= −340 N·m

81

4.1.6GOAL: Determine the work needed to lift a block.GIVEN: Weight of block and change in height.DRAW:

FORMULATE EQUATIONS: The work done has a positive component (due to F ) and anegative component (due to gravity). The total work done is found from W

1−2=∫ 6 ft0

(F −mg)dx

and our energy equation is

KE∣∣∣1+ W

1−2= KE

∣∣∣2

(1)

SOLVE: The speed at state 1 and 2 is zero and therefore KE∣∣∣1

= KE∣∣∣2

= 0Both forces acting on the block are constant and thus their work integral is simply the force timesthe displacement:

(1) ⇒ Fh−mgh = 0 ⇒ Fh = mgh

Work done by F = mgh = (50 lb)(6 ft) = 300 lb· ft

82

4.1.7GOAL: Find required constant force to project rock to given height.GIVEN: Desired height and parameters.DRAW:


Rectilinear motion with constant acceleration: v23− v2

2= 2a(y

3− y

2) (1)

Work/energy:

∫ y2

y1

Fdy = W1−2 = KE∣∣∣2−KE

∣∣∣1

(2)

SOLVE:(1) ⇒ (0 m/s)2 − v2

2= −2g(10 m− 0.5 m) (3)

(3) ⇒ v22

= 2(9.81 m/s2)(9.5 m) (4)

(2) ⇒0.5m∫0

(F −mg)dy = (F − (5 kg)(9.81 m/s2))(0.5 m) =12(5 kg)2(9.81 m/s2)(9.5 m) (5)

(5) ⇒ F = 981N

83

4.1.8GOAL: Determine the coefficient dynamic friction needed to bring a sliding mass to rest afterrising 5 m.GIVEN: Mass, slope and initial speed.DRAW:


FORMULATE EQUATIONS: The FBD gives the forces on the mass as

N⇀b2 − µN

⇀b1 −mg⇀

In addition to this, we’ll use our work/energy formulation:12

mv21

+ W1−2 =12

mv22

SOLVE: The forces we need be concerned with (the ones that act along the path) are given by

F = −µN −mg sin θ

A force balance in the⇀b2 direction gives us N = mg cos θ and so our applied force becomes

F = −mg(µ cos θ + sin θ)

Applying work/energy from state 1 to state 2 gives us12

mv21−mg(µ cos θ + sin θ)d = 0

µ =v21− 2g sin θd

2g cos θd=

(10.2 m/s)2 − 2(9.81 m/s2)(0.5)(10m)2(9.81 m/s2)(0.866)(10 m)

µ = 0.035

84


⇀ı ⇀⇀b1 1/

√2 1/

√2

⇀b2 −1/

√2 1/

√2


N⇀b2 + µN

⇀b1 −mg⇀

In addition to this, we’ll use our work/energy formulation:12

mv21

+ W1−2 =12

mv22

SOLVE: The forces we need be concerned with (the ones that act along the path) are given by

F = mg sin θ − µN

A force balance in the⇀b2 direction gives us N = mg/

√2 and so, using θ = 45◦, our downslope

force becomes

F =mg(1− µ)√

2First let’s consider the case of µ = 0. In this case we have

0 +mgd√

2=

12

mv22

(9.81 m/s2)(10m)√2

=v22

2⇒ v

2= 11.8 m/s

Next, we’ll let µ = 0.1:

0 +mgd(1− µ)√

2=

12

mv22

(9.81 m/s2)(10m)(1.0− 0.1)√2

=12

v22⇒ v

2= 11.2 m/s

The mass is moving 0.604 m/s slower due to the friction, a 5.1% decrease.

85



(−mg + a + bx)⇀

To solve the problem we’ll use the energy/work formula:12

mv21

+ W1−2 =12

mv22

SOLVE: First let’s find the contact speed. We’re given that the pinecone falls 30 m and so have

mgh =12

mv21

v1

=√

2gh =√

2(9.81 m/s2)(30m) = 24.3 m/s

The force along the path of travel is given by

F = mg − a− bx

and to determine the work done we’ll simply integrate with respect to displacement.12

mv21

+ mgd−∫ d

0(a + bx)dx = 0

x denotes distance traveled beneath the snow surface and d distance traveled when the pineconefinally comes to rest.

12

mv21

+ mgd− ad− bd2

2= 0

bd2

2+ (a−mg)d− 1

2mv2

1= 0

d2 +2(a−mg)

bd−

mv21

b= 0

d2 +2[4N− (0.3 kg)(9.81 m/s2)]

14 N/md− (0.3 kg)(24.3 m/s)2

14 N/m= 0

d = 3.48 m

86

4.1.11GOAL: Determine the launch speed of a payload from a catapult.GIVEN: Force acting along the direction of travel, size of the catapult and the initial and finalangle of the catapult arm.DRAW:

⇀ı ⇀⇀er − cos θ sin θ⇀eθ sin θ cos θ


Fθ

⇀eθ + N ⇀er −mg⇀

To solve the problem we’ll use the energy/work formula:12

mv21

+ W1−2 =12

mv22

SOLVE:First we need the total force F that acts along the direction of travel:

F = Fθ−mg cos θ

Our work/energy expression becomes

0 +∫ π

2

π4

(Fθ−mg cos θ)rdθ =

12

mv22

Fθrπ

4−mgr(sin

π

2− sin

π

4) =

12

mv22

(10 N)(1.5 m)π

4− (0.45 kg)(9.81 m/s2)(1.5 m)(sin

π

2− sin

π

4) =

12

(0.45 kg)v22

v2

= 6.61 m/s

87

4.1.12GOAL: Determine the average force acting on a bullet as it travels through a rifle barrel.GIVEN: Length of the barrel, mass of the bullet and exit speed.DRAW:

FORMULATE EQUATIONS:To solve the problem we’ll use the energy/work formula:

12

mv21

+ W1−2 =12

mv22

SOLVE:

700 mph = (700 mph)(

88 ft/s60 mph

)= 1027 ft/s

Our work/energy expression is given by

F∆x =12

mv22

F (4 ft) =12

2 oz

(16 oz/lb)(32.2 ft/s2

) (1027 ft/s)2

F = 511 lb

88

4.1.13GOAL: Find the final speed of block B.GIVEN: µs = µ

d= 0.6, m

A= 10 kg, m

B= 20 kg, y

2− y

1= 0.5 m

DRAW

FORMULATE EQUATIONS: The force balance on block A is⇀ı : m

Ax = T − F

A

⇀ : 0 = N −mA

g

where the force of sliding friction is given by FA

= µN . Thus, the force balance can be resolvedinto a single equation:

mA

x = T − µmA

g

The force balance on block B ism

By = m

Bg − T

Lastly, the energy balance on the system is

KEA

∣∣∣2+KE

B

∣∣∣2

= KEA

∣∣∣1+KE

B

∣∣∣1+ W

A

∣∣∣1−2

+ WB

∣∣∣1−2

(1)

ASSUME: Let’s assume that the rope is inelastic, which yields the following constraints:

y2− y

1= x

2− x

1⇒ y = x (2)

SOLVE: The work done on block A is

WA

∣∣∣1−2

=∫ x

2

x1

(T − µmA

g)dx = (T − µmA

g)(x2− x

1)

The work done on block B is

WB

∣∣∣1−2

=∫ y

2

y1

(mB

g − T )dx = (mB

g − T )(y2− y

1)

Using the constraint, the total work done on the system becomes

W∣∣∣1−2

= (T − µmA

g + mB

g − T )(y2− y

1) = g(m

B− µm

A)(y

2− y

1)

89

Recognizing that the kinetic energy of the system at State 1 is zero, (1) can be written as

12

mA

x2 +12

mB

y2 = g(mB− µm

A)(y

2− y

1) (3)

(2) → (3) ⇒ y =

√√√√2g

(m

B− µm

A

mA

+ mB

)(y

2− y

1)

y =

√2(9.81 m/s2)

[20 kg− (0.6)(10 kg)

10 kg + 20 kg

](0.5 m) = 2.14 m/s

y = 2.14 m/s

90

4.1.14GOAL: Find the speed of A when the friction force is equal end opposite of the insertion force.GIVEN: m = 2 kg, F = 12 N, Fs = k(x

0+ αx), k = 100 N/m, x

0= 0.05 m, α = 0.27, µ = 0.4,

starts from rest at x = 0DRAW:

ASSUME: The spring model only serves to model the force change during the insertion and isn’tviewed as itself requiring work to accomplish the compression. We only consider the work done byfriction and by the 12N insertion force. The spring/wall interface is frictionless.FORMULATE EQUATIONS:We’ll apply work/energy:

KE∣∣∣2

= KE∣∣∣1+ W

∣∣∣1−2

KE∣∣∣1

= 0 ⇒ KE∣∣∣2

= W∣∣∣1−2

=∫ x

f

0Fxdx

FBD=IRD N ⇀ −mg⇀ − Fs⇀ + F ⇀ı − F

f⇀ı = mx⇀ı

SOLVE:⇀ : N −mg − Fs = 0 ⇒ N = mg + Fs = mg + k(x

0+ αx)

Ff

= µN = µ(mg + k(x

0+ αx)

)In order to find W

∣∣∣1−2

we need to find xf. The goal is to find the speed of A when the friction

force is equal and opposite of the insertion force of 12 N. Find xf

that satisfies this condition.

F = µ(mg + k(x

0+ αx

f))

xf

=

((F

µ −mg)

k− x

0

)(1α

)=

12 N0.4 − (2 kg)(9.81 m/s2)

100 N/m− 0.05 m

( 10.27

)= 0.2 m

W∣∣∣1−2

=∫ x

f

0Fxdx =

∫ xf

0F − F

fdx

F − Ff

= F − µ(mg + k(x

0+ αx)

)= A + Bx

where A and B are chosen to simplify the expressions.

A = F−µ(mg+kx0) = 12 N−(0.4)

[(2 kg)(9.81 m/s2) + (100 N/m)(0.05 m)

]= 2.15 N

B = −µkα = −(0.4)(100N/m)(0.27) = −10.8 N/m

91

W∣∣∣1−2

=∫ x

f

0(A + Bx)dx =

[Ax +

Bx2

2

] ∣∣∣xf

0

W∣∣∣1−2

= Axf

+Bx2

f

2= (2.15 N)(0.2 m) +

(−10.8 N/m)(0.2 m)2

2= 0.214 J

KE∣∣∣2

= W∣∣∣1−2

= 0.214 J =12mv2 ⇒ v = 0.463 m/s

⇀v = 0.463⇀ı m/s

92

4.1.15GOAL: Find the force acting against a car as it brakes to a stop.GIVEN: Car weighs=2800 lb and goes from 60 mph to zero in 143 ft.DRAW:

FORMULATE EQUATIONS:We’ll apply work/energy:

KE∣∣∣2

= KE∣∣∣1+ W

∣∣∣1−2

FBD:∑ ⇀

F = N ⇀ −mg⇀ − Fb

⇀ı

SOLVE:

m =2800 lb

32.2 ft/s2= 87.0 slg

Applying work/energy gives us

0 =12

mv2 +143 ft∫0

−Fbdx

Fb

=12 (87.0 slg)(88 ft/s)2

143 ft

Fb

= 2.35×103 lb

93

4.1.16GOAL: Determine the work done on an aluminum pellet as it travels through an imaginary gun’sbarrel. Express the work in an equivalent manner to illustrate its size.GIVEN: Length of the barrel, mass of the pellet and exit speed.DRAW:


12

mv21

+ W1−2 =12

mv22

SOLVE:The exit speed is given as one quarter of the speed of light. We’ll ignore relativistic effects in thefollowing calculations.

v1

= 0, v2

=3.00×108 m/s

4= 7.49×107 m/s

Our work/energy expression is given by

W1−2 =12

mv22

W1−2 =12

(0.005 kg)(7.49×107 m/s)2 = 1.40×1013 N·m

Now let’s compare this to the work associated with lifting a 1500 kg car in a 1 g gravitational field.The work done will simply be equal to mgh where h is the height that the car is lifted.

(1500 kg)(9.81 m/s2)h = 1.40×1013 N·m

h =1.40×1013 N·m

(1500 kg)(9.81 m/s2)= 9.54×108 m

Considering the fact that the distance from the earth to the moon is about 3.84×108 m, this meansthat the energy to shoot the gun is equal to that needed to move a normally sized car more thantwice as far as from the earth to the moon. What does that tell you about the probability thatsomeone could actually stand in one place while firing such a weapon?

94

4.1.17GOAL: Determine the maximum impact speed of a vehicle such that it comes to rest just as itreaches the end of a deformable barrier.GIVEN: Force profile as a function of deflection and weight of the vehicle.DRAW:

FORMULATE EQUATIONS: To solve the problem we’ll use the energy/work formula:12

mv21

+ W1−2 =12

mv22

SOLVE: The force along the path of travel is given by

F = −a− bx12

and to determine the work done we’ll simply integrate with respect to displacement.12

mv21−∫ d

0(a + bx

12 )dx = 0

x denotes distance traveled after contacting the barrier and d is the distance traveled when the carfinally comes to rest.

12

mv21− ad− 2

3bd

32 = 0

v =

√√√√√2[ad +

23bd

32

]m

v =

√√√√√√√√2[(5000 lb)(30 ft) +

23(3000 lb/ ft

12 )(30 ft)

32

]3800 lb

32.2 ft/s2

v2

= 90.1 m/s

95

4.1.18GOAL: Find compression of a forklift’s bumper.GIVEN: Bumper force characteristics, initial speed and forklift massDRAW:


KE∣∣∣2

= KE∣∣∣1+ W

1−2

SOLVE:

KE∣∣∣1

=12

mv21

=12

(1000 kg)(1.5 m/s)2 = 1125N·m

KE∣∣∣2

= 0

W1−2

= −x∫

0

(2.7×107 N/m2)x2dx =−(2.7×107 N/m2)x3

3

∣∣∣∣∣x

0

= −(9×106 N/m2)x3

0 = 1125 N·m− (9×106 N/m)x3 ⇒ x = 0.050 m

96

4.1.19GOAL: Find the speed of a block after a compressed spring has releasedGIVEN: Mass, spring constant, compressionDRAW:

FORMULATE EQUATIONS: State 1 is with the spring fully compressed and mass stationary.State 2 is with the spring completely uncompressed. There are no nonconservative work terms.

KE1

+ PE1

= KE2

+ PE2

SOLVE:

KE1

= 0; PESP1

=12kx2 =

12

(981 N/m) (0.08 m)2 = 3.14 N·m; PEg1

= 0

KE2

=12mx2 = (0.1 kg) x2

PESP2

= 0

PEg2

= mgh = (0.2 kg)(9.81 m/s2

)(0.08 m) = 0.157 N·m

3.14 N·m = (0.1 kg) x2 + 0.157 N·m

x = 5.46 m/s

97

4.1.20GOAL: Determine the tension in a pair of restraining strings and the height above the floorattained by the released block.GIVEN: Spring characteristics and mass of block.DRAW:


12

mv21

+ W1−2 =12

mv22

SOLVE:We initially have a static force balance in the vertical direction:

−2T cos 30◦ −mg + 2k∆x = 0

T =2(500 N/m)(0.05 m)−mg

2 cos 30◦= 50 N−mg√

3

We see from this that as the weight of the block (mg) increases, the tension decreases and, for asufficiently heavy block, the strings would go slack.Now we move onto the case when the strings break. We’ll consider three total states. State 1 iswith the block at its lowest point (0.01 m above the floor with the spring compressed 0.05 m). State2 is when the spring has reached its fully extended state (block is 0.06 m above the floor with aspeed to be determined). After state 2 the block is in a free trajectory and will reach a maximumheight when gravity has decreased its speed to zero (state 3).State 1 to state 2:

2k(∆x)2

2−mg(0.05 m) =

12

mv22

(1)

From state 2 to state 3 we have12

mv22

= mg∆y (2)

(1), (2) ⇒2k(∆x)2

2−mg(0.05 m) = mg∆y

∆y = −0.05 m +k(∆x)2

2mg

The total height h is the height at state 2 (0.06 m) plus the change in height from state 2 to state3 (∆y). Thus we have

98

h = 0.06 m + ∆y = 0.01 m +2k(∆x)2

2mg= 0.01 m +

(500 N)(0.05 m)2

2m(9.81 m/s2)

h = 0.01 m + 0.127 kg·mm

Clearly the mass of the block will alter the ultimate height attained (as expected from physicalconsiderations). Note that the analysis assumes that the mass isn’t so great as to prevent the springfrom completely extending. The way to determine if this assumption is valid is to simply evaluateh for a given value of m. If h exceeds 0.06 m, the unstretched spring length, then we know that mwas “small enough” to match our assumption.

99

4.1.21GOAL: Determine the speed of a mass particle at |θ| = 30◦ along a circular path.GIVEN: Particle’s mass, shape of path, initial velocity and position.DRAW:


12

mv21

+ W1−2 =12

mv22

SOLVE:We’ll first determine if the mass has enough energy to reach the top of the hill with a finite speed. Ifso, we know that it will then move to the right side of the hill (θ negative) and thus will eventuallyreach θ = −30◦. If it turns out not to have enough energy to reach θ = 0 then the conclusion isthat it stops somewhere partway up, reverses direction and eventually reaches θ = 30◦. The forcedue to gravity that acts against the mass along its trajectory is mg sin θ.From state 1 to state 2 we have

12

mv21

+∫ 0

θ0

mg sin θrdθ =12

mv22

v22

= 2[12

(1.25 m/s)2 − (9.81 m/s2)(1m)(1− cos 20◦)]

v22

= 0.379( m/s)2

v22

is positive, implying a real solution. Our conclusion is that it does reach θ = 0 and thereforewill pass θ = −30◦.Denoting its position at θ = 30◦ as state 3 we have

12

mv21

+∫ −30◦

θ0

mg sin θrdθ =12

mv23

v23

= 2[12

(1.25 m/s)2 − (9.81 m/s2)(1m)(cos(−30◦)− cos 20◦)]

⇀vm = −1.73⇀eθ m/s

100

4.1.22GOAL: Determine the height that a sliding mass particle will obtain.GIVEN: Friction force, mass of particle and initial speed and position.DRAW:


12

mv21

+ W1−2 =12

mv22

SOLVE:12

mv21−∫ θ

0

0F

frdθ −

∫ θ0

0mg sin θrdθ = 0

12

mv21− F

frθ

0+ mgr(cos θ

0− 1) = 0

(0.05 kg)(1.2 m/s)2−(0.6 N)(1.2 m)θ0+(0.1 kg)(9.81 m/s2)(1.2 m)(cos θ

0−1) = 0

θ0

= 0.0929 rad

101

4.1.23GOAL: Determine the speed of a mass particle at |θ| = 30◦ along a circular path.GIVEN: Particle’s mass, shape of path, initial velocity and position.DRAW:

FORMULATE EQUATIONS:To solve the problem we’ll use the energy/work formula, applied twice.

12

mv21

+ W1−2 =12

mv22

SOLVE:We’ll first determine what the speed of the mass is when it reaches state B, starting from state C.Next, we’ll consider the change in speed to to work done from state B to the final state A.From state C to state B we have

12

mv2C

+∫ 35◦

0mg sin θhdθ =

12

mv2B

Note that Ff

= 0 in this phase.Using v

C= 0 gives us

v2B

= 2 [−hg(cos 35◦ − 1)]

From geometry we have d = h/ tan 35◦. Thus

h = d tan 35◦ = 7mand

v2B

= 24.85( m/s)2

Now we go from state B to A at a constant slope of 35◦. The normal force N is seen to be equal tomg cos 35◦ and thus we know that a resisting force of µmg cos 35◦ acts to oppose m’s motion downthe slope.

12

mv2B

+ mg sin 35◦d− µmg cos 35◦d =12

mv2A

v2A

= 24.85( m/s)2 + 2gd(sin 35◦ − µ cos 35◦)

v2A

= 24.85( m/s)2 + 80.39( m/s)2 = 105.2( m/s)2

⇀vA = 10.3⇀eθ m/s

102

4.1.24GOAL: Determine the speed of a skier at the end of a downhill run.GIVEN: Skier’s mass, shape of path, initial velocity and position.DRAW:

FORMULATE EQUATIONS:To solve the problem we’ll use the energy/work formula, applied twice, once during the straightfrictional phase from A to B and then again along the curved, friction-free phase from B to C.

12

mv21

+ W1−2 =12

mv22

SOLVE:We’ll first determine what the speed of the mass is when it reaches state B, starting from state A.Next, we’ll consider the change in speed to to work done from state B to the final state C.From state A to state B the frictional force opposing the acceleration due to gravity is equal toµ

dN = µ

dmg2 .

0 +∫ 60m

0

(√3mg

2−

µdmg

2

)ds =

12

mv2B

(9.81 m/s2)

(√3

2− 0.08

2

)(60 m) =

12

mv2B

vB

= 31.2 m/s

Now we go from state B to C with a friction-free interface. We need to integrate along the circulararc, accounting for how the component of force due to gravity changes as the position along thearc changes.

12

mv2B

+ (100 m)∫ π

2rad

π3rad

mg cos θdθ =12

mv2C

12

(32.2 m/s)2 + (100 m)(9.81 m/s2)(

sin(

π

2rad

)− sin

(π

3rad

))=

12

v2C

vC

= 44.2 m/s

103

4.1.25GOAL:Find the velocity of the mass when x = −5 mGIVEN: x = −5.0 m/s, µ = 0.4, m = 3kgDRAW:


N⇀b2 + S

⇀b1 −mg⇀ = m⇀a

⇀a = x⇀b1

N⇀b2 + S

⇀b1 −mg(sin 20◦

⇀b1 + cos 20◦

⇀b2) = mx

⇀b1

⇀b1: mx = S −mg sin 20◦ (1)

⇀b2: 0 = N −mg cos 20◦ (2)

(2)⇒ N = mg cos 20◦ (3)

If the mass is slipping , then S = µN .Thus:

mx = µmg cos 20◦ −mg sin 20◦ = mg(µcos 20◦ − sin 20◦)

The force acting on the mass in the x direction is mg(µcos 20◦ − sin 20◦).STATE 1:

x = 0mSTATE 2:

x = −5 m

KE∣∣∣1+ W

1−2= KE

∣∣∣2

12m(5.0 m/s)2 −mg(µcos 20◦ − sin 20◦)(5m) = KE

∣∣∣2

=12mv2

2

12(5.0 m/s)2 − (9.81 m/s2) [(0.4)(0.9397)− 0.3420] (5 m) =

12v22

v2

= 4.66 m/s

104

4.1.26GOAL:Analyze a particle in two different reference framesFORMULATE EQUATIONS:(a) YESInertial reference frames can differ by a constant translational velocity. A stationary particle wouldappear to move (equal and opposite) when viewed from a moving frame. Thus the kinetic energywould differ.(b) YESFor the same reason as for (a). The distance ”traveled” by the particle will vary along with itsperceived velocity.(c) YESThe changes in (a) and (b) are consistent and the correct work/energy relations will hold.Example:Frame 1 is stationary. Frame 2 moves to the left at v

0m/s. A constant force F pushes a mass

particle m to the right for t seconds.Frame 1:

Since⇀F = m⇀a, we have ⇀a =

⇀Fm .

∆x = at2

2=

Ft2

2m

W1−2

= F∆x =F 2t2

2m

KE∣∣∣1+ W

1−2= KE

∣∣∣2

0 +F 2t2

2m= KE

∣∣∣2

⇒ KE∣∣∣2

=F 2t2

2m(1)

Check:v after t seconds is given by:

v = at =Ft

m

KE∣∣∣2

=12mv2 =

12m

F 2t2

m2=

F 2t2

2m(2)

(1) and (2) match, as expected.Now look at the system from frame 2:Initially the mass is stationary. Since the frame is moving left at v

0, the perceived velocity is v

0(to the right).

KE∣∣∣1

=12mv2

0

In t seconds the mass is perceived to move its actual change in displacement plus the amount theframe has moved:

∆x = at2

2+ v

0t =

Ft2

2m+ v

0t

W1−2

= F∆x =F 2t2

2m+ Fv

0t

105

KE∣∣∣2

= KE∣∣∣1+ W

1−2=

12mv2

0+

F 2t2

2m+ Fv

0t (3)

Check:v after t seconds is given by:

v = at + v0

=Ft

m+ v

0

KE∣∣∣2

=12m

(Ft

m+ v

0

)2

=12

(F 2t2

m2+

2v0Ft

m+ v2

0

)

KE∣∣∣2

=F 2t2

2m+ Fv

0t +

12mv2

0(4)

(3) and (4) agree, showing that work/energy holds.(d) FALSEWork/energy was derived from Newton’s second law.

106

4.2 Potential Energies and Conservative Forces

107

4.2.1GOAL: Find v when compression is 0.8 m. Discuss.GIVEN: System parameters and target compression amount.DRAW:

FORMULATE EQUATIONS:We’ll apply conservation of energy

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

SOLVE:

0 + mgh =12

mv2 +12

kx2

(5 kg)(9.81 m/s2)(1m) =12

(5 kg)v2 +12

(400 N/m)(0.8 m)2

v2 = −31.6(m/s)2 ⇒ v is imaginary

b) Imaginary velocities aren’t physically possible. Thus mathematics are telling us that the springcan’t compress .8 m. The maximal compression must be less than 0.8 m.Check: Find x when v = 0

mgh =12

kx2

(5 kg)(9.81 m/s2)(1m) =12

(400 N/m)x2 ⇒ x = 0.5 m < 0.8 m!

108

4.2.2GOAL: Find θ for loss of contact.DRAW:

⇀ı ⇀⇀er − sin θ − cos θ⇀eθ − cos θ sin θ


Force balance: −mg⇀ −N ⇀er = m[−rθ2 ⇀er + rθ⇀eθ] (1)

Our second equation involving θ and θ comes from conservation of energy:

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

(2)

SOLVE: For loss of contact we’d want N = 0 and thus

rθ2

= −g cos θ (3)

(2) ⇒12

mv21

=12

m(rθ)2 + mgr(1− cos θ) (4)

(3) → (4) ⇒12

mv21

= −12

mrg cos θ + mgr(1− cos θ) (5)

12

mv21

= mgr(1− 32

cos θ) (6)

12

(4 kg)(7.95 m/s)2 = (4 kg)(9.81 m/s2)(1.5 m)(1− 32

cos θ)

cos θ = −0.765

θ = 140◦

109

4.2.3GOAL: Find the distance travelled by the mass.GIVEN: k = 100N/m, m

A= 0.05 kg, µs = 0.6, µ

d= 0.3, initial spring compression of 0.1 m

DRAW

⇀ı ⇀⇀b1 cos 20◦ sin 20◦⇀b2 − sin 20◦ cos 20◦

FORMULATE EQUATIONS: Let’s first check if the force of static friction is sufficient to holdthe mass against the compressed spring.

Force balance: mA

s⇀b1 = F

spring

⇀b1 − F

fmax

⇀b1 + N

⇀b2 −m

Ag⇀

where Ffmax

= µsN , Fspring

= kx, and x is the compression of the spring.

SOLVE: Looking just at the⇀b1 component of the force balance gives us

mA

s = kx−mA

g(µ cos 20◦ + sin 20◦)

mA

s = (100N/m)(0.1 m)− (0.05 kg)(9.81 m/s2)(0.6 cos 20◦ + sin 20◦) = 9.56 N

Thus, there is a positive force in the⇀b1 direction even when the force of static friction is at its

maximum. So we can now solve the problem knowing that dynamic friction is acting on the mass.FORMULATE EQUATIONS: Let’s now look at the case in which the mass has started at restwhile pushed up against the spring, and ends up a distance d further up the ramp. Because themass starts at rest, and ends at rest, the energy balance is

PE∣∣∣2

= PE∣∣∣1+ Wnc

∣∣∣1−2

(1)

SOLVE: The non-conservative work done by friction is:

Wnc

∣∣∣1−2

=∫ s

2

s1

Ffds =

∫ s2

s1

(−µdm

Ag cos 20◦)ds

Wnc

∣∣∣1−2

= −µdm

Ag(s

2− s

1) cos 20◦ = −µ

dm

Agd cos 20◦

The expressions for potential energy at the two states are

PE∣∣∣1

= mA

gs1sin 20◦ +

12

kx2

PE∣∣∣2

= mB

gs2sin 20◦

Plugging these into (1) gives us

mA

gd sin 20◦ =12

kx2 − µdm

Agd cos 20◦

110

Solving for d yields

d =12 kx2

mA

g(sin 20◦ + µd

cos 20◦)=

0.5(100 N/m)(0.1 m)2

(0.05 kg)(9.81 m/s2) [sin 20◦ + (0.3) cos 20◦]= 1.63 m

d = 1.63 m

111

4.2.4GOAL: Find the cushion’s maximum compression.GIVEN: m = 0.2 kg, h = 1.5 m, k = 40N/mDRAW:

ASSUME: The track is frictionless.FORMULATE EQUATIONS:We’ll apply conservation of energy:

KE∣∣∣2+ PE

∣∣∣2

= KE∣∣∣1+ PE

∣∣∣1

PE∣∣∣2

= PE∣∣∣1

SOLVE:0.5kx2 = mgh

x =

√2mgh

k=

√2(0.2 kg)(9.81 m/s2)(1.5 m)

(40 N/m)= 0.384 m

x = 0.384 m

112

4.2.5GOAL: Two methods to impart velocity to a mass are proposed. Evaluate them and determine ifone will produce a higher velocity.GIVEN: Starting and ending configuration of the two methods.DRAW:

FORMULATE EQUATIONS:All we need is to apply conservation of energy:

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

SOLVE:Case A:

0 + mgh =12

my2 + 0

y =√

2gh

⇀v impact = −√

2gh ⇀

Case B:The speed at impact will be given by hθ.

0 + mgh =12

m(hθ)2

+ 0

hθ =√

2gh

⇀v impact = −hθ ⇀ı = −√

2gh ⇀ı

CHECK: The impact velocities are identical and therefore both methods will crack coconutsequally well. From a design perspective I would have to say Case B would make for more repeatableand accurate strikes but Case A has the advantage of no moving parts so I would have to call it atie.

113

4.2.6GOAL: Find spring constant to limit spring compression to specified amount.GIVEN: Initial speed and parameters. L = 5m, m = 2kg, θ = 30◦, v = 8m/s.DRAW:

FORMULATE EQUATIONS: We’ll apply conservation of energy:

KE∣∣∣2+ PE

∣∣∣2

= KE∣∣∣1+ PE

∣∣∣1

(1)

SOLVE:

(1) ⇒ 0 +12k(0.1 m)2 + mg(L + 0.1 m) sin θ =

12mv2 (2)

(2) ⇒ k = (2 kg)((8 m/s)2 − (9.81 m/s2)(5.1 m))0.1 m2 = 2.79×103 N/m

114

4.2.7GOAL: Determine the impact speed of a falling cyclist.GIVEN: Cyclist’s initial orientation and distance from the ground.DRAW:

FORMULATE EQUATIONS:We will employ conservation of energy from state 1 to state 2:

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

SOLVE:We’re told that the cyclist simply topples over and thus at the point of impact with the ground ismoving straight down (from geometry). Our energy conservation equation becomes

0 + mgh =12

mv2 + 0

The mass drops out and we’re left with

(9.81 m/s2)(1m) =12

v2

v = 4.43 m/s

115

4.2.8GOAL: Determine the impact speed of a mass against the ceilingGIVEN: Initial configuration of the system and spring constants.DRAW:

FORMULATE EQUATIONS:We will employ conservation of energy from State 1 to State 2:

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

SOLVE:Initially (State 1) the kinetic energy is zero and the springs are stretched by an amount

√42 + 22 m−

1 m and at State 2 are stretched (2m− 1 m). Our energy conservation equation is

212

k(√

42 + 22 m− 1 m)2

=12

mv2 + 212

k(2 m− 1 m)2 + mg(4 m)

212

(80 N/m) (3.472 m)2 =12

(10 kg)v2+212

(80 N/m)(1m)2+(10 kg)(9.81 m/s2)(4m)

v = 9.92 m/s

116

4.2.9GOAL: Determine the maximum compression of a spring that models the elastic elements of a leg.GIVEN: Initial configuration of the system, mass and spring constant.DRAW:

FORMULATE EQUATIONS:We will employ conservation of energy from state 1 to state 2:

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

SOLVE:At State 1 the spring has not yet compressed. At State 2 the speed is zero. Our conservation ofenergy equation is

12

mv2 =12

kh2 −mgh

where h is the compression of the leg spring.12

(10 kg)(2 m/s)2 =12

(14, 000 N/m)h2 − (10 kg)(9.81 m/s2)h

h = 6.09 cm

117

4.2.10GOAL: Determine the maximum height off the ground a rebounding toy will reach.GIVEN: Initial configuration of the system, mass and spring constant.DRAW:


KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

SOLVE:At State 1 the spring is fully compressed and the mass has zero speed. At State 2 the speed isagain zero. Our conservation of energy equation is

12

k(0.04 m− 0.008 m)2 = mg(h− 0.008 m)

12

(3000 N/m)(0.04 m− 0.008 m)2 = (0.08 kg)(9.81 m/s2)(h− 0.008 m)

h = 1.97 m

118

4.2.11GOAL: Determine the effective spring constant of a trampoline.GIVEN: Maximum deflection of the trampoline as a result of a falling body and the mass of thebody.DRAW:

FORMULATE EQUATIONS:We will employ conservation of energy from State 1 to State 2, State 2 to 3, and State 3 to 4.SOLVE:We’re given that the upward speed at State 1 is 12 ft/s. After reaching a maximum height, themass will come back down and, arriving at the same height at State 2, will have the same speedbut oriented down instead of up. Thus v

2= v

1= 12 ft/s. We can apply conservation of energy

from State 2 to 3 and then from 3 to 4 in a single step. We’ll define the zero potential energy stateas that associated with the mass at the level of the undeformed trampoline, as in State 3.

12

m(12 ft/s)2 + mg(50 ft) = −mg(1.3 m) +12

k(1.3 ft)2

12

(150 lb

32.2 ft/s2

)(12 ft/s)2 + (150 lb)(50 ft) = −(150 lb)(1.3 m) +

12

k(1.3 ft)2

k = 9503 lb/ft

119

4.2.12GOAL: Find the speed with which a mass particle hits an elastically mounted stop after rebound-ing once.GIVEN: System parameter values.DRAW:


FORMULATE EQUATIONS:We’ll use work/energy

KE∣∣∣1+ PE

∣∣∣1+ W

1−2= KE

∣∣∣2+ PE

∣∣∣2

and break the analysis into segments.SOLVE:Initially we have a potential energy of mg(5 m) and zero kinetic energy. Negative work is done asthe mass drops due to friction, with the work equal to −µNd = −µmgd cos θ (where N = mg cos θis the normal force between the mass and slope and d is the distance moved along the slope).At the end of this first phase of motion the mass has dropped a distance h below it’s zero point(the vertical position of the undeformed spring pad. Our energy balance is therefore

0 + mg(5 m)− µmg cos θ

(5 m + h

sin θ

)= −mgh +

12

k

(h

sin θ

)2

Using θ = 40◦ and solving with the given parameter values (realizing that k = 150, 000 N/m) yieldsh = 1.015×10−2 m. This is quite a small deflection and reflects the fact that the spring basicallyacts to rebound the mass without much “give” of its own.Thus at State 2 we have zero speed, the mass has moved slightly below its original zero potentialenergy position and we have a negative gravitational potential energy and a positive spring potentialenergy. Next we calculate how high the mass will reach upon rebound:

12

k

(h

sin θ

)2

−mgh− µmg

(h + y

3

tan θ

)= mgy

3

Solving this for the given parameter values and using h = 1.015×10−2 m yields a rebound heightof y

3= 3.071 m.

Our final calculation equates the potential energy at a height y3

minus the loss due to the finalslide down the incline with the kinetic energy at the spring pad:

mgy3−mgµ

(y3

tan θ

)=

12

mv2

v = 6.94 m/s

120

4.2.13GOAL: Find the speed with which hinge B is moving after the system has moved under theinfluence of a stretched spring.GIVEN: Initial and final system configuration, link lengths, masses and spring constant.DRAW:


FORMULATE EQUATIONS:We’ll use work/energy for this no-loss system

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

The challenge here is to correctly determine the speed of both B and C, which we’ll do by use ofthe law of sines:

x

sin γ=

BC

sin θ

where the angles and sides are illustrated in State 2 of the figure.SOLVE:At State 1 we have a total energy of

E = mB

gh+12

k(1.2 m−0.2 m)2 = (10 kg)(9.81 m/s2)(0.4 m)+12

(80 N/m)(1m)2 = 79.24 N·m (1)

At State 2, C is moving to the right at a speed x and B is rotating about A with a speed |vB| =

(0.5 m)|θ|.x

sin γ=

BC

sin θ⇒ x sin θ = BC sin γ = BC sin(180◦ − θ − β)

Differentiating with respect to t gives us

x sin θ + xθ cos θ = −BC cos(180− θ − β)(θ + β) (2)

To go further we need to find β in terms of θ. Looking at State 2 in the figure gives us (fromgeometry)

(0.5 m) sin θ = (0.4 m) sinβ (3)

and

121

x = (0.5 m) cos θ + (0.4 m) cos β (4)

Differentiating (3) yields

(0.4 m)β cos β = (0.5 m)θ cos θ

β =(

(0.5 m) cos θ

(0.4 m) cos β

)θ (5)

Using (3) with θ = 30◦ gives us β = 38.68◦ and (4) yields x = 0.745 m.Using these values of x and β, along with (5), and substituting them into (2) gives us

θ = −(1.676 m−1)x

We can now evaluate the energy at State 2. C has moved to the right and the total stretch of thespring is now (1.2 m− (0.745 m− 0.3 m)− 0.2 m) = 0.555 m.The energy when θ = 30◦ is

E =12

k(0.555 m)2 + mB

g(0.25 m) +12

mC

v2C

+12

mB

v2B

E =12

(80 N/m)(0.555 m)2+(10 kg)(9.81 m/s2)(0.25 m)+12

(15 kg)x2+12

(10 kg)[(0.5 m)(−1.676 m−1)x]2

E = 36.83 N·m + (11.01 kg)x2 (6)

(1), (6) ⇒ x = 1.96 m/s

|vB| = |(0.5 m)θ| = |(0.5 m)(−1.676 m−1)(1.96 m/s)|

|vB| = 1.64 m/s

122

4.2.14GOAL: Find if collar reaches B and, if so, its speed at B.GIVEN: Mass of collar, spring constant, system geometry and initial speedDRAW:


KE∣∣∣1+ PE

∣∣∣g1

+ PE∣∣∣sp

1

= KE∣∣∣2+ PE

∣∣∣g2

+ PE∣∣∣sp

2

SOLVE:

KE∣∣∣1

=12

mv21

= 0

PE∣∣∣g1

= 0

PE∣∣∣sp

1

=12

kx21

=12

(30 N/m)(0.7 m− 0.1 m)2 = 5.4 N·m

KE∣∣∣2

=12

mv22

= (0.25 kg)v22

PE∣∣∣g2

= mgh = (0.5 kg)(9.81 m/s2)(0.4 m) = 1.962 N·m

PE∣∣∣sp

2

=12

kx22

=12

(30 N/m)(0.4 m− 0.1 m)2 = 1.35 N·m

(0.25 kg)v22

= (5.4− 1.962− 1.35)N·m

v2

= 2.89 m/s

Clearly, the mass reaches B, as shown by the real solution for v2.

123

4.2.15GOAL: Determine the minimum spring constant k so that a mass does not impact the bottom ofa dropped enclosure.GIVEN: Size of enclosure, number of springs and arrangement, mass and clearance in enclosure.DRAW:


PE∣∣∣1+KE

∣∣∣1

= PE∣∣∣2+KE

∣∣∣2

SOLVE:We’re given that the speed of the enclosure and mass just before impact is 8 m/s. Thus the masswill still have this speed even though the enclosure is brought to an abrupt stop. At State 1 themass has a finite kinetic energy and also a finite potential energy due to its height above the ground.At State 2 we assume zero speed, zero potential energy due to gravity and a potential energy dueto the extension/compression of the positioning springs.

PE∣∣∣1+KE

∣∣∣1

= PE∣∣∣2+KE

∣∣∣2

mgh +12

mv2 =12

(2k)h2

(0.02 kg)(9.81 m/s2)(0.04 m) +12

(0.02 kg)(8 m/s)2 =12

(2k)(0.04 m)2

k = 404.9 N/m

124

4.2.16GOAL: Find v

Bafter |θ| = 45◦

GIVEN: geometry and massesDRAW:

FORMULATE EQUATIONS: To determine rotation, perform a moment balance about O∑M

O= m

Bgr −m

AgL sin 30◦ (1)

followed by conservation of energy:

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

(2)

SOLVE:

(1) ⇒ mB

gr−mA

gL sin 30◦ =(9.81 m/s2

)[(0.5 kg) (0.08 m)− (1.1 kg) (0.6 m) (0.5)] = −2.84 N·m

The moment sum is negative and thus the disk rotates clockwise

KE∣∣∣1

= 0

PEg

∣∣∣1

= mA

gL cos 30◦ = (1.1 kg)(9.81 m/s2

)(0.6 m) (0.866) = 5.61 N·m

KE∣∣∣2

=12L2θ2m

A+

12r2θ2m

B

PEg

∣∣∣2

= mB

gr

(π

4rad− π

6rad

)+ m

AgL cos 45◦

= (0.5 kg)(9.81 m/s2

)(0.08 m)

(π

4rad− π

6rad

)+ (1.1 kg)

(9.81 m/s2

)(0.6 m)

(1√2

)= 4.68 N·m

(2) ⇒ 5.61 N·m = 4.68 N·m +12θ2[(0.6 m)2 (1.1 kg) + (0.08 m)2 (0.5 kg)

]θ = 2.15 rad/s

125

⇀vB = rθ⇀ = (0.08 m) (2.15 rad/s) ⇀ = 0.172⇀ m/s

126

4.2.17GOAL: Determine the speed of a mass (part of a two-mass/spring system) when the orientationof the supporting link has moved from its initial position to a new one.GIVEN: Inclination angle is initially 60◦ and finally 45◦. Mass and spring constants as well assystem geometry.DRAW:


KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

SOLVE:Initially (State 1) the kinetic energy is zero and the spring is stretched by an amount 2(1.2 m) sin 60◦−1.23 m = 0.848 m. At State 2 the spring is stretched by 2(1.2 m) sin 45◦ − 1.23 m = 0.467 m. Bothmasses will be moving at the same speed (from symmetry) and we’ll call this speed v.Our energy conservation equation is

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

12

k (0.848 m)2+2mg(1.2 m) cos 60◦ =12

(2m)v2+12

k (0.467 m)2+2mg(1.2 m) cos 45◦

12

(42 N/m) (0.848 m)2 + 2(2.0 kg)(9.81 m/s2)(1.2 m) cos 60◦ =

12

(2)(2.0 kg)v2 +12

(42 N/m) (0.467 m)2 + 2(2.0 kg)(9.81 m/s2)(1.2 m) cos 45◦

v = 0.626 m/s

From geometry we see that B’s velocity is oriented up and to the left:⇀vB = 0.626(− 1√

2⇀ı + 1√

2⇀ı ) m/s = 0.443(−⇀ı + ⇀ ) m/s

127

4.2.18GOAL: Determine the speed of a mass (part of a mass/two spring system) when it has droppeda set distance beneath its release position.GIVEN: Mass and spring constants, initial and final system orientation.DRAW:


KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

SOLVE:Initially (State 1) the kinetic energy is zero and the springs are unstretched. At State 2 each springis stretched by

√0.012 + 0.042 m− 0.01 m = 0.03123 m. The mass will be moving at speed v.

Our energy conservation equation is

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

mg(0.04 m) =12

mv2 + 2(

12

)k (0.03123 m)2

(0.5 kg)(9.81 m/s2)(0.04 m) =12

(0.5 kg)v2 + 2(

12

)(60 N/m) (0.03123 m)2

v = 0.742 m/s

We know the mass is dropping down and thus have⇀vB = −0.742⇀ m/s

128

4.2.19GOAL: Find the distance the cyclist can coast.GIVEN: v

1= 25mph, 6% grade, no air resistance or friction

DRAW


FORMULATE EQUATIONS: The energy balance is given by

KE∣∣∣2+ PE

∣∣∣2

= KE∣∣∣1+ PE

∣∣∣1

(1)

SOLVE:

Rearranging (1) ⇒ PE∣∣∣2− PE

∣∣∣1

= KE∣∣∣1−KE

∣∣∣2

(2)

Because the cyclist will be at rest at state 2, KE∣∣∣2

= 0. The change in potential energy in terms ofd is given by

PE∣∣∣2− PE

∣∣∣1

= mgd sin θ (3)

(3) → (2) ⇒ mgd sin θ =12

mv21

d =v21

2g sin θ=

(25 mph× 1.4667 ft/s

mph

)2

2(32.2 ft/s2) sin [tan−1(0.06)]= 348.6 ft

d = 348.6 ft

129

4.2.20GOAL: Find the speed at which the separated block hits the ceiling.GIVEN: m = 10 kg, k = 200N/mDRAW:

ASSUME: The spring has an unstretched length of zero and will not provide a force when theblock hits the ceiling.FORMULATE EQUATIONS:We’ll apply conservation of energy:

KE∣∣∣2+ PE

∣∣∣2

= KE∣∣∣1+ PE

∣∣∣1

KE∣∣∣2+ PE

∣∣∣2

= PE∣∣∣1

KE∣∣∣2

= PE∣∣∣1− PE

∣∣∣2

SOLVE:First use equilibrium to find yeq , the initial stretched length of the spring.

FBD=0 kyeq⇀ −mg⇀ = 0 ⇒ yeq =

mg

k

PE∣∣∣1

=12k(yeq )

2 =k

2(mg

k)2 =

m2g2

2k

PE∣∣∣2

=m

2gh =

mg

2(yeq ) =

mg

2(mg

k) =

m2g2

2k

KE∣∣∣2

= PE∣∣∣1− PE

∣∣∣2

=m2g2

2k− m2g2

2k= 0

KE∣∣∣2

= 0 ⇒ v = 0

v = 0

130

4.2.21GOAL: Find speed when fishing pole is straight.GIVEN: Initial conditions and parameters.DRAW:

FORMULATE EQUATIONS: We’ll apply conservation of energy:

KE∣∣∣2+ PE

∣∣∣2

= KE∣∣∣1+ PE

∣∣∣1

(1)

An equilibrium force balance before the fish drops off gives us

ky − (mA

+ mB

)g = 0 (2)

where y is the distance that the end of the rod has sagged under a gravitational load.SOLVE:

(2) ⇒ k =(m

A+ m

B)g

y(3)

(1), (3) ⇒12m

Av2 =

12ky2 −m

Agy (4)

(3), (4) ⇒ 12(0.04 kg)v2 =

12

(0.04 kg + 10 kg)(9.81 m/s2)0.2 m

(0.2 m)2 − (0.04 kg)(9.81 m/s2)(0.2 m)

v = 22.1 m/s

131

4.2.22GOAL: Determine the variation in speed of a sliding block under friction-free vs friction conditions.GIVEN: Angle of sloped surface, coefficient of friction. The mass slides 2 m along the slope.DRAW:

⇀ı ⇀⇀b1 cos 15◦ sin 15◦⇀b2 − sin 15◦ cos 15◦

FORMULATE EQUATIONS: In both cases we’ll utilize work/energy including both potentialand kinetic energy terms:

KE∣∣∣1+ PE

∣∣∣1+ W

1−2= KE

∣∣∣2+ PE

∣∣∣2

(1)

SOLVE:

First we’ll look at the zero friction case. Using (1) with no included dissipative work term, alongwith a height change of h

1, gives us

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

mgh1

=12

(5 kg)s2

(5 kg)(9.81 m/s2)(2 sin 15◦m) =12

(5 kg)s2

s = ±3.19 m/s

We know from the geometry of the problem that the mass is moving downslope when it reachesState 1 and so we have

⇀vB = −3.19⇀b1 m/s

132

Now consider a coefficient of friction µ = 0.2.

Force balance: ms⇀b1 = F

⇀b1 + N

⇀b2 −mg⇀

Force balance: ms⇀b1 = F

⇀b1 + N

⇀b2 −mg(sin 15◦

⇀b1 + cos 15◦

⇀b2)

⇀b1 : ms

⇀b1 = F

⇀b1 −mg sin 15◦ (2)

⇀b2 : 0 = N −mg cos 15◦ ⇒ N = (5 kg)(9.81 m/s2) cos 15◦ = 47.38 N (3)

First let’s verify that the mass does, indeed, slip downslope. If we assume that s = 0 then (2)implies

F = mg sin 15◦ = 12.7 N

Thus, the force needed to keep the mass from accelerating downslope is 12.7 N. The maximumfrictional force obtainable is given by

Fmax = µN = 0.2(47.38 N) = 9.48 N

The maximum frictional force is less than that needed to hold the mass stationary and thereforewe have slip.

(1) ⇒ KE∣∣∣1+ PE

∣∣∣1+ W

1−2= KE

∣∣∣2+ PE

∣∣∣2

mgh1− |F |(2 m) =

12

ms2

(5 kg)(9.81 m/s2)(2m sin 15◦)− (9.48 N)(2 m) =12

(5 kg)s2

s = ±1.60 m/s

Again, we know the mass moves downslope and therefore have⇀vB = −1.60

⇀b1 m/s

The addition of the friction slowed the mass by

100(

3.19−1.603.19

)= 0.50, i.e. by 50 percent

133

4.2.23GOAL: Determine if a trampoline bottoms out when a person jumps into it and, if so, how highthe supports must be raised to avoid bottoming.GIVEN: Height that person jumps from (20 ft), spring constant of trampoline (40 lb/in) andperson’s weight (122 lb).DRAW:


m =122 lb

32.2 ft/s2= 3.79 slg, k = (40 lb/in)(12 in/lb) = 480 lb/ft

We’ll approach the problem in two steps. First we’ll find the speed of impact with the undeformedtrampoline (a fall of 17 ft). Then we’ll assume a zero velocity state (State 3) associated with adeflection h of the trampoline. If h is less than 3 ft then we can conclude the trampoline doesn’tbottom out. If h is more than 3 ft then we immediately have the required height.SOLVE:Assume that when at State 2 the falling mass has zero velocity. Because we start with zero velocityas well, there is no kinetic energy term in our energy balance.

PE∣∣∣1+KE

∣∣∣1

= PE∣∣∣2+KE

∣∣∣2

mg(17 ft) =12

mv2

(122 lb)(17 ft) =12

(3.79 slg)v2

v2 = 1095( ft/s)2

Now we proceed from State 2 to State 3:

PE∣∣∣2+KE

∣∣∣2

= PE∣∣∣3+KE

∣∣∣3

mgh +12

mv2 =12

kh2

12

(3.79 slg)(1095( ft/s)2) + (3.79 slg)(32.2 ft/s2)h =12

(4880 lb/ft)h2

h2 − (0.508 ft)h− 8.65 ft2 = 0

h = 3.19 ftSince 3.19 ft > 3 ft we see that the answer to part (a) is yes - the trampoline will bottom out. The

answer to (b) is that to avoid bottoming out, the trampoline must be at least 3.19 ft above theground.

134

4.2.24GOAL:Verify the numerical results from Problem 3.5.28 using energy methods. Find the coefficient offriction which causes the sliding mass to stop at θ = 30◦.GIVEN: r = 20m, θ = 0.5 rad/s and µ

d= 0.1.

The following is a program to calculate the kinetic energy of the mass particle as work is done to itfrom friction and gravity. The plots show the exact kinetic energy (1

2 mv2) versus the kinetic energycalculated from a work standpoint. As you can see, the results are in agreement.

135

136

4.2.25GOAL: Determine the velocity of a mass particle after traveling through a tube.GIVEN: The size and shape of the tube, mass of the particle and initial velocity.DRAW:

FORMULATE EQUATIONS:We’ll use both energy conservation

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

SOLVE:Because the tube’s surface is frictionless, it doesn’t matter that the mass is traveling along somethingother than a straight, vertical path - the kinetic energy change is simple due to the change inpotential energy from the start to the finish. All forces generated between the mass and the wallsof the tube are normal to the direction of travel and hence do no work.

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

12

mv21

=12

mv22

+ mg(2 m)

12

(0.1 kg)(11 m/s)2 =12

(0.1 kg)v22

+ (0.1 kg)(9.81 m/s2)(2m)

⇀v2 = 9.04⇀ m/s

137

4.2.26GOAL: Determine if an actor can successfully swing across a chasm.GIVEN: The breaking strength of the vine is 1100 N and the actor has a mass of 80 kg. The totalincludes angle of the vine from start to finish is 60◦.DRAW:

FORMULATE EQUATIONS:We’ll use both energy conservation

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

and a force balance at the bottom of the swing

(T −mg)⇀ = m(rθ2 ⇀ + rθ⇀ı )

SOLVE:⇀ı : 0 = mrθ

⇀ : T −mg = mrθ2 ⇒ T = mg + mrθ2 (1)

Now let’s apply conservation of energy.

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

12

mv21

+ mgh1

=12

mv22

+ mgh2

Using h1

= (10 m)(1− cos 30◦) = 1.34 m, and h2

= 0 we have

(80 kg)(9.81 m/s2)(1.34 m) =12

(80 kg)v22⇒ |v

2| =

v2

= 5.127 m/s

v22

= [(10 m)θ]2 ⇒ θ2 = 0.2629 ( rad/s)2 (2)

(1), (2) ⇒ T = (80 kg)(9.81 m/s2) + (80 kg)(10 m)(0.2629( rad/s)2) = 995 N

Because the vine’s tensile strength is 1,100 N, more than the needed 995 N, the stunt can besuccessfully completed.

138

4.2.27GOAL: Find the velocity of Block A when mass B strikes the ground.GIVEN: Bar is massless and has length L, torsional spring applies M = −k

θθ

mA

= 5kg, mB

= 10 kg, L = 0.8 m, kθ

= 10 kg·m/radDRAW:

FORMULATE EQUATIONS:State 1:Pendulum uprightState 1:Pendulum horizontal

KE∣∣∣1

= 0 (1)

PE∣∣∣1

= mB

gL +12k

θ

(π

2

)2

(2)

KE∣∣∣2

=12m

Av2A

+12m

Bv2B

(3)

PE∣∣∣2

= mA

g(sin(30◦))∆xA

(4)

2∆xA

= ∆y (5)

where ∆xA

is the motion of block A up the slope and ∆y is the distance the rope going over thepulley moves. The figure entitled Labeling shows that the length of rope from the pulley C to massB is initially 0.5L and finally

√(1L)2 + (1.5L)2 = 1.8L.

∆length of rope = 1.80L− 0.5L = 1.30L

Therefore:

139

∆y = 1.30L = 2∆xA

∆xA

= 0.65L (6)

To determine the kinetic energy of block A we need to have its speed. We know from kinematicsthat its speed is half that of the rope going over the pulley C. This rope ultimately attaches tomass B, which at State 2 is moving vertically downward. We can break v

Bat State 2 into two

components, aligned along⇀b1 and

⇀b2:

We can see from geometry that|y| = |v

B| sinβ

Thus we have|v

B| sinβ = 2|v

A| (7)

SOLVE:

(7)→(3)⇒ KE∣∣∣2

=12m

A(12vB

sinβ)2 +12m

Bv2B

(8)

(6)→(4)⇒ PE∣∣∣2

=m

Ag

2(0.65L) (9)

(1),(2),(8),(9)⇒ mB

gL +12k

θ

(π

2rad

)2

=18m

Av2B

sin2 β +12m

Bv2B

+ mA

g(0.325)L

(10 kg)(9.81 m/s2)(0.8 m) +12(10 N·m/rad)

(π

2rad

)2

=

v2B

((5 kg)

8(sin(33.7))2 +

(10 kg)2

)+ (5 kg)(9.81 m/s2)(0.325)(0.8 m)

5.192 v2B

= 78.1 m2/s2

vB

= 3.88 m/s

140

4.2.28GOAL: Find safe length of bungie cord and impact velocity if cord is too weak.GIVEN: Initial and final height of jumper, mass of jumper and spring constant of bungie cord.DRAW:

FORMULATE EQUATIONS: Conservation of energy

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

(1)

SOLVE:(a) Kinetic and potential energies

KE∣∣∣1

= 0, PEg

∣∣∣1

= mg(70 m), PEbc

∣∣∣1

= 0 (2)

KE∣∣∣2

= 0, PEg

∣∣∣2

= mg(3 m), PEbc

∣∣∣2

= 12 k(67 m− L)2 (3)

Using (1) we get

(55 kg)(9.81 m/s2)(70m) = (55 kg)(9.81 m/s2)(3m) +12

(22 N/m)(67m− L)2 (4)

L = 9.67 m

(b) Kinetic and potential energies

KE∣∣∣1

= 0, PEg

∣∣∣1

= mg(70 m), PEbc

∣∣∣1

= 0 (5)

KE∣∣∣2

= 12 mv2, PEg

∣∣∣2

= 0, PEbc

∣∣∣2

= 12 k(70 m− L)2 (6)

Again, using (1) we get

(55 kg)(9.81 m/s2)(70m) =12

(55 kg)v2 +12

(0.9)(22N/m)(70m− 9.67 m)2 (7)

v = 7.95 m/s

141

4.2.29GOAL: Find the maximum deflection of the spring after a moving mass strikes it.GIVEN: System parameter values.DRAW:

⇀ı ⇀⇀b1 sinβ − cos β⇀b2 cos β sinβ

FORMULATE EQUATIONS: A force balance gives us

N⇀b2 − S

⇀b1 −mg⇀ = mx

⇀b1

If the mass slides then S = µN . Resolving the equations of motion into the⇀b1 and

⇀b2 directions

yields:

mg cos β − µN = mx

N −mg sinβ = 0

Which, when combined, give us:

mx = mg(cos β − µ sinβ)

Thus, the force acting on the mass along the direction of the travel is mg(cos β − µ sinβ).Work-energy gives us

PE∣∣∣1+KE

∣∣∣1+ W1−2 = PE

∣∣∣2+KE

∣∣∣2

SOLVE: At top of the slide

KE∣∣∣1

= 0, PEg

∣∣∣1

= mgL cos β

At the instant of contact with the spring we have

KE∣∣∣2

=12

mv22, PEg

∣∣∣2

= −mgL cos β, W1−2 = −µNL

Invoking the work-energy between these two states

−µmgL sinβ =12

mv22 −mgL cos β

−0.1(9.81 m/s2)(2m)12

=12

v22 − (9.81 m/s2)(2m)

√3

2

142

v2 = 5.66 m/s. At full compression we have

KE∣∣∣3

= 0, PEg

∣∣∣3

= 0, PEs

∣∣∣3

=12

k(∆x)2

Using work-energy between contact and full compression gives

12

mv22 − µmg sin β∆x = −mg cos β∆x +

12

k(∆x)2

12

(1.1 kg)v22 − 0.1(1.1 kg)(9.81 m/s2)

12∆x = −(1.1 kg)(9.81 m/s2)

√3

2∆x +

12

(3500 N/m)(∆x)2

(∆x)2 − (5.032× 10−3 m)(∆x)− 1.006× 10−2 m2 = 0

∆x = 0.1029 m

143

4.2.30GOAL:Determine θ as a function of r and other constants.Determine r as a function of r and other constants.Assuming L = 0, what is rmax for the trajectory?Verify part (c) by numerically integrating the equation of motion.GIVEN: System configuration and parameter values.DRAW:

(a):


Apply conservation of angular momentum about O:

Angular momentum about O at A: H0

= mvD (1)

Angular momentum at arbitrary position: H0

= mr(rθ) = mr2θ (2)

ASSUME: Because the spring force acts in a purely radial direction it won’t affect theangular momentum and thus we have

(1), (2) ⇒ mr2θ = mvD

SOLVE:

θ = vDr2 (3)

(b):


Apply conservation of energy:

12

mv2 +12

k(D − L)2 =12

k(r − L)2 +12

m

((rθ)2

+ r2)

(4)

SOLVE:

(3) → (4) ⇒ mv2 + k(D − L)2 = k(r − L)2 + m

(D2v2

r2+ r2

)

r =√

v2(r2−D2)r2 + k

m (D2 − 2DL + 2rL− r2)

(c):

FORMULATE EQUATIONS: rmax occurs when drdt = 0. Setting r = 0 and L = 0 in the

preceding equation yields:

144

v2(r2 −D2

)r2

+k

m

(D2 − r2

)= 0

v2

r2=

k

m

r2 =v2m

k

rmax = v√

mk = 1.483 m

(d):

FORMULATE EQUATIONS: The equations of motion are found from F = ma:

m[(

r − rθ2)

⇀er +(rθ + 2rθ

)⇀eθ

]= −k(r − L)⇀er

⇀er: m(r − rθ2) = −k(r − L) (5)

⇀eθ: rθ + 2rθ = 0 (6)

ASSUME: There’s no need to integrate (6) because from conservation of angular momentumabout O, we have

θ =vD

r2(7)

SOLVE:

(7)→(5)⇒ r = − k

m(r − L) +

(vD)2

r3(8)

Shown below is a segment of the data output from MATLAB after integrating the precedingequation. Note that at t = 0.2332 s we have r = 1.4832 m, a precise match to the theoreticalprediction. The m-file code and a graphical plot of r(t) vs t is also shown.

t r(t) r(t)0.1995 1.4498 1.97950.2062 1.4619 1.58740.2130 1.4713 1.19090.2197 1.4779 0.79100.2265 1.4819 0.38890.2332 1.4832 −0.01430.2400 1.4817 −0.41740.2467 1.4776 −0.81940.2535 1.4707 −1.21910.2602 1.4611 −1.6154

145

function dy=wk7p6(t,y)y1=y(1);y2=y(2);dy(1,1)=y(2);dy(2,1)=-50*y1/1.1+25/y1^3;

146

4.2.31GOAL:a) Find the value of µ for which the mass will not stay in contact with the hoopGIVEN: m has an initial speed of vm = 3

√gr = 9.905 m/s

DRAW:


Force balance: m((r − rθ2)⇀er + (rθ + 2rθ)⇀eθ

)= −mg⇀ −N ⇀er − S⇀eθ

m((r − rθ2)⇀er + (rθ + 2rθ)⇀eθ

)= ⇀er(mg cos θ−N)+⇀eθ(−mg sin θ−S)

⇀er: m(r − rθ2) = mg cos θ −N

⇀eθ: m(rθ + 2rθ) = −mg sin θ − S

GEOMETRIC CONSTRAINTS:r = r = 0

m(−rθ2) = mg cos θ −N

mrθ = −mg sin θ − S

If the mass is moving in the positive θ direction, then S = µN . Hence we have:

−mrθ2 = mg cos θ −N (1)

mrθ = −mg sin θ − µN (2)

(1)⇒ N = mg cos θ + mrθ2 (3)

(1),(3)⇒ mrθ = −mg sin θ − µ(mg cos θ + mrθ2)

θ + µθ2 +g

r(µ cos θ + sin θ) = 0 (4)

(4) is our equation of motion and (3) lets us determine the normal force between the mass and thehoop.(a)The minimum θ value at θ = π rad can be found from (3) by setting N to zero:

147

0 = mg(−1) + mrθ2min

θmin

=√

g

r

To find the corresponding θ value at θ = 0, we can apply conservation of energy:STATE 1:

θ = π rad, θ =√

g

r

STATE 2:θ = 0

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

2rmg +12m(rg) = 0 +

12m(rθ)2

5rg = r2θ2

θ =√

5gr at θ = 0

NUMERICAL:(b)For this part, we need to find the µ such that at θ = π, the normal force N goes to zero. θ at θ = 0is set to 9.905 rad/s. By using MATLAB and numerically integrating for a variety of µ values it asfound that µ = 0.1555 causes the loss of contact to occur at θ = π.

148

4.2.32GOAL: Find the maximum angle θ0 the boy can swing.GIVEN: System parameter values.DRAW:

⇀ı ⇀⇀er − sin θ − cos θ⇀eθ − cos θ sin θ

ASSUME: The rope is inextensible and therefore its length is constant. Thus r = L and r and rare zero.FORMULATE EQUATIONS: A force balance gives us

−T ⇀er −mg⇀ = m(−Lθ2 ⇀er + Lθ⇀eθ)

(mg cos θ − T )⇀er −mg sin θ⇀eθ = m(−Lθ2 ⇀er + Lθ⇀eθ)

The tension of the string does no work (it is perpendicular to the path). Therefore, conservationof energy holds

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

(1)

SOLVE:The maximum tension in the rope occurs at he bottom of the swing, where the potential energy isminimized and therefore the kinetic energy is at a maximum. Looking at the force balance in the

⇀er and ⇀eθ directions for θ = 0 gives us

θ = 0 (2)

T = mLθ2 + mg (3)

Identifying State 1 as the system inclined at an angle θ and State 2 as the mass hanging straightdown we have

KE∣∣∣1

= 0, PEg

∣∣∣1

= mgL(1− cos θ) (4)

KE∣∣∣2

= 12 m(Lθ2)2, PEg

∣∣∣2

= 0 (5)

149

Substituting in (1)

mgL(1− cos θ) =12

m(Lθ2)2 (6)

We know that Tmax = (40 kg + 25 kg)(9.81 m/s2) = 637.7 N at θ = 0. Using (3) we get

637.7 N = (40 kg)(4 m)θ22

+ (40 kg)(9.81 m/s2) ⇒ θ22

= 1.53 s−2 (7)

Substitute in (6) and solving for the associated θ gives us

θ1

= 46.6◦

150

4.2.33GOAL:Find the maximum height, force on the center of mass, and speed at liftoff.GIVEN: You weigh 80 lb, you can jump 1 foot in a stationary elevator, your center of mass moves6 inches, the elevator’s speed is 18 in/s.FORMULATE EQUATIONS:Stationary elevator:STATE 1:on the groundSTATE 2:at the top of the leap

PE∣∣∣1+KE

∣∣∣1

= PE∣∣∣2+KE

∣∣∣2

0 +12

(80 lb

32.2 ft/s2

)v2 =

(80 lb

32.2 ft/s2

)(32.2 ft/s2)(1 ft) + 0

v = 8 ft/s

To attain a speed of 8 ft/s your legs muscles had to push against the force due to gravity.Assume your center of mass shifted 1

2 ft during the leap. Redefining our states as:STATE 1:Beginning of leg extensionSTATE 2:End of leg extension

PE∣∣∣1+KE

∣∣∣1+ W

1−2= PE

∣∣∣2+KE

∣∣∣2

0+0+F12

=

(80 lb

32.2 ft/s2

)(32.2 ft/s2)

(12

)+

12

(80 lb

32.2 ft/s2

)(8 ft/s)2

F = 239.0 lbYour legs produced 239 lb of force to launch you into your leap.Decelerating elevator:Now consider the leg extension and subsequent leap in a decelerating elevator. During the legextension phase, we have:

m(y + z) = F −mg(80 lb

32.2 ft/s2

)(y − 12 ft/s2) = 239 lb− 80 lb

y(t) = 76 ft/s2

y(t) = (76 ft/s2)t

y(t) = (76 ft/s2)t2

2

y(t∗) =12

= (76 ft/s2)(t∗)2

2

⇒ t∗ = 0.115 s

151

y(t∗) = 76 ft/s2(0.115 s) = 8.72 ft/s

Your speed at the start of your leap is greater (8.72 ft/s) than in the case of a stationary elevator(8 ft/s). The difference between your speed and the decelerating elevator is given by:

s = (18 ft/s + 8.72 ft/s)− (32.2 ft/s2)t− (18 ft/s− (12 ft/s)t)

s = 8.72 ft/s− (20.2 ft/s2)t

This equals zero (maximum separation) at :

t = 8.72 ft/s20.2 ft/s2

= 0.432 s

152

4.2.34GOAL: Find the normal force between the skater and the track.GIVEN: System configuration and parameters.DRAW:

FORMULATE EQUATIONS: State 1 corresponds to the skater stationary at a height of 16 ftabove the ground and State 2 corresponds to the skater at a height for which θ = 45◦.To determine the normal force, we need to know the skate boarder’s speed. We can find this froman energy approach.

h2(t) = 10 ft(1− 1√

2) = 2.929 ft

At State 1 all the energy is due to the gravitational potential:

E = mgh1

= (70 lb)(16 ft) = 1120 lb·ft

At State 2 we have both kinetic and gravitational energy:

E = mgh2

+12mv2 = (70 lb)(2.929 ft) +

35 lb32.2 ft/s2

v2

Equating the energy at States 1 and 2 gives:

1120 lb·ft = 205 lb·ft +35 lb

32.2 ft/s2v2

v = 29 ft/s

Now that we have the velocity, we can determine the forces:

Force Balance: m(rθ⇀eθ − rθ2 ⇀er) = −mg⇀ −N ⇀er

⇀er: −mrθ2 = mg sin θ −N

N =70 lb

32.2 ft/s2(32.2 ft/s2

(1√2

)+

(29 ft/s)2

10 ft)

N = 232.5 lb

This is quite a bit of normal force. The skateboarder has to withstand over 3g’s at the halfwaypoint down the curve.

153

4.2.35GOAL: Analyze a bead/track interface.GIVEN: r = 0.4 m, each bead has a speed of 0.01 m/s.ASSUME: No friction between beads and wire.DRAW:

FORMULATE EQUATIONS:We know that at the midpoints, E and F , both beads must be moving at the same speed (fromenergy conservation) and in the same direction (both wires are oriented vertically). Thus the onlyvariation can occur in the top half.For the right wire we have:

y = g

y = gt + v0

y =gt2

2+ v

0t

We are given y(0) = 0.01 m/s and so have:

y(t) = (9.81 m/s2)t + 0.01 m/s (1)

y(t) = (4.905 m/s2)t2 + (0.01 m/s)t (2)

For the left wire we have:Force Balance: −mg⇀ + N ⇀er = m(rθ⇀eθ − rθ2 ⇀er)

⇀eθ: mrθ = mg sin θ

θ =g

rsin θ (3)

We can determine when the bead on the right reaches the midpoint from:

r =gt2

2+ (0.01 m/s)t

154

t2 +0.02 m/s9.81 m/s2

t− 2(0.4 m)9.81 m/s2

= 0

t2 + (2.04×10−3 s)t− 8.15×10−2 s2 = 0

t = 0.285 s

We have to numerically integrate (3) to find the time for the bead to reach θ = π2 rad. The initial

conditions are θ(0) = 0, θ(0) = 0.01m/s0.4m = 0.025rad/s. Using ode45 in MATLAB yields a time of :

t = 1.3098 s

This is quite a bit longer than was the case for a purely vertical drop. Because the speed entry isthe same for both beads at E and F , their behavior is identical from that point on.This analysis has shown that it takes a different amount of time for the bead to reach the samevertical position for the two cases. Now we’ll look at the bead’s speed. Because the kinetic energyfor both cases will vary purely dependent upon the bead’s vertical position (the potential energychanges with vertical position and there is no friction to affect the energy balance) we would expectidentical speeds when the bead’s are at the same vertical position. We can use the numericalintegration results for the left wire and use the analytical results for the right one.A plot of v versus y for each case shows identical speed versus vertical position characteristics -the two plots fall on top of each other. The velocities differ, due to the different orientation of thewire, but the speeds (and hence the kinetic energies) are the same.

155

4.2.36GOAL: Show that KE

2= KE

1+ W

1−2“works”.

GIVEN: Form of the governing equation and the numerical approximation to use.FORMULATE EQUATIONS:

(0.5 kg)x = sinπt N (1)

SOLVE:Shown at the end of this solution are the two M-files used in this problem. wk7p20.m just integrates(1). The output is a vector of times t and states y. They’re combined in the workspace into d: d[t,y].After checking on the number of data points (n), n and d are used as inputs to wk7p20a.m. Thiscomputes the kinetic energies, which are plotted below. In the first plot, for which the work is givenby F (tn)(x(t

n+1)−x(tn)), there is somewhat reasonable agreement but a noticeable error near the

end. This is because, even though the force is changing from x(tn) to x(tn+1

), the approximationtreats it as fixed.

Letting the work be given byF (tn )+F (t

n+1)

2 [x(tn+1

)−x(tn)] corrects this error by using the averageF over each time interval. The corresponding plot is almost identical to the plot of 1

2mx2.

156

157

4.2.37GOAL: Find the initial and final kinetic energy of the mass.GIVEN: Physical geometry and parameter values of the cockroach model.DRAW:

FORMULATE EQUATIONS:Force balance,mass:

−mg ⇀ − k(r − L)⇀er − cr ⇀er = m((r − rθ2)⇀er + (rθ + 2rθ)⇀eθ

)(1)

⇀er: −mg sin θ − k(r − L)− cr = m(r − rθ2) (2)

⇀eθ: −mg cos θ = m(rθ + 2rθ) (3)

Kinetic Energy, mass: KE =12

mv2 (4)

SOLVE:

(2)→ (3)⇒ d

dt

rrθ

θ

=

r

−g sin(θ)− k(r−L)m − cr

m + rθ2

θ

−gr cos(θ)− 2rθ

r

(5)

The initial conditions are

r(0) = L (6)r(0) = v

0cos(θ0) (7)

θ(0) =π

3(8)

θ(0) =−r0 tan(θ0)

r0(9)

Integrating using MATLAB (see code), the final speed is 0.1723 m/s.At θ = π

3 , KE = 5× 10−5J , and at θ = 2π3 , KE = 3.71× 10−5J

%Main Fileglobal g L c m k

%Giveng = 9.8; % m/s^2L = .005; % mc = 1; % N*s/mm = 0.0025; % kg

158

k = 4000; % N/mv0 = -0.2; % m/s to the lefttfinal = 1; %seconds

%Calculate initial conditionsr0 = L; % Spring initially unstretchedtheta0 = pi/3; % Initial touchdown anglerdot0 = v0/ (cos(theta0)+tan(theta0)*sin(theta0)); % Initial radial velocity m/sthetadot0 = -rdot0/r0*tan(theta0); % Initial angular velocity rad/s

[t,y]=ode45(@BAM,[0 tfinal],[r0 rdot0 theta0 thetadot0]);r = y(:,1);rdot = y(:,2);theta = y(:,3);thetadot = y(:,4);

stop = max(find(theta<2*pi/3)); %Find when angle = 2*pi/3;polar(theta(1:stop),r(1:stop));vf = sqrt(rdot(stop)^2+ (r(stop)*thetadot(stop))^2);

KEi = 1/2*m*v0^2KEf = 1/2*m*vf^2

%------------------------------Separate File: BAM.m -------------------function xdot = BAM(t, x);global g L c m k %x(1) = r, x(2) = rdot, x(3) = theta, x(4) = thetadot

% Differential Equationxdot(1) = x(2);xdot(3) = x(4);xdot(2) = -g*sin(x(3)) - k/m*(abs(x(1))-L)-c/m*x(2)+x(1)*x(4)^2;xdot(4) = -g/x(1)*cos(x(3)) - 2*x(2)*x(4)/x(1);xdot = [xdot(1); xdot(2); xdot(3); xdot(4)];

159

4.2.38GOAL: Find the maximum height reached and the maximum compression of the pogo stick.GIVEN: Simplified pogo model and its parameter values.DRAW:


State 1: KE∣∣∣1

= 12 mv2 PE

∣∣∣1

= mgh0

(1)


= 0 PE∣∣∣2

= mgh2

(2)


= 0 PE∣∣∣3

= mgh3

+12

k|h3 − L|2 (3)

ASSUME:

Total energy is conserved:

E = KE + PE = Constant (4)

SOLVE:

(1),(2),(4)⇒12

mv2 + mgh0

= mgh2

h2

=v2

2g+ h

0=

(2.5 m/s)2

2(9.81 m/s2)+ 1.5 m

h2

= 1.8 m

(1),(3)⇒12

mv2 + mgh0

= mgh3

+12

k|h3− L|2 (5)

Numerically solving⇒ h3

= 0.558 m (6)

|h3− L| = 0.642 m

160

4.2.39GOAL: Find angular velocity of the bowling ball at its lowest point.GIVEN: Geometry of arm/bowling ball model and parameters.DRAW:


KE∣∣∣1+ PE

∣∣∣1+ W1→2 = KE

∣∣∣2+ PE

∣∣∣2

0 + mgL +

θf∫0

M dθ =12

mv2 + mgL(1− sin θ) (1)

GIVEN:

θf

= π/2

SOLVE:

(1)⇒ vf

=

√2(

gL +Mπ

2m

)

161

4.2.40GOAL: Find the needed spring constant to allow Tarzan to reach Jane. Plot the Tarzan’s swingpath up to reaching Jane.GIVEN:m

1= 81 kg, m

2= 60 kg, d = 20m, h = 12m, L = 15m.

DRAW:

ASSUME: Tarzan’s velocity vector should be oriented horizontally when reaching Jane’s position.GOVERNING EQUATIONS:Conservation of energy:

PE∣∣∣1+KE

∣∣∣1

= PE∣∣∣2+KE

∣∣∣2

SOLVE: Applying our energy balance gives us

mTg(12 m) +

12

k(1 m)2 + 0 =12

mTv2 +

12

k(6 m)2

(81 kg)(9.81 m/s2)(12m) +12

k(1 m)2 + 0 =12

(81 kg)v2 +12

k(6 m)2

9.535×103 N·m = (40.5 kg)v2 + k(17.5 m2)

Note that this is one equation in two unknowns. There exist an infinite number of solutions.Physically, they correspond to Tarzan at the level of the quicksand with a velocity vector that isn’tnecessarily horizontal. If we confine ourselves to an energy approach, we won’t be able to determinethe particular k for which Tarzan contacts Jane while moving horizontally. To solve the problemwe need to consider the system kinetics.

162

⇀ı ⇀⇀er − cos θ − sin θ⇀eθ sin θ − cos θ

The FBD=IRD diagram is shown along with a polar coordinate system.

Force balance: mT(r − rθ2) + m

T(rθ + 2rθ) = −T ⇀er −m

Tg⇀

⇀er : mT(r − rθ2) = −T + m

Tg sin θ (1)

⇀eθ : rθ + 2rθ = g cos θ (2)

(1) ⇒ r = g sin θ − k(r − 14 m)m

T

+ rθ2 (3)

(2) ⇒ θ =g cos θ − 2rθ

r(4)

We can put these equations into state form and integrate in MATLAB, using initial conditions ofr = 15m, r = 0, θ = 0.562 rad and θ = 0. The output data can be used to determine the x, ycoordinates of Tarzan, using O (the attachment of the vine) as an origin:

x = −r cos θ, y = 20m− r sin θ

By plotting the results, looking to see where the low point of the trajectory occurs and then alteringk and iterating, we can ultimately find

k = 292N/m

A plot of the trajectory is shown below.

163

4.2.41GOAL: Determine the spring constant k needed to limit the compression of a restraining block to5 cm. Determine as well how much a 1400 kg car would settle under a gravitation load if it hadequivalent springs.GIVEN: m

b= 9000 kg and v = 0.3 m/s.

DRAW:

FORMULATE EQUATIONS:STATE 1:Just before collision, v = 0.3 m/s and the spring is uncompressed.STATE 2:At the maximum spring compression the boat has zero speed.The general energy conservation equation is

PE∣∣∣1+KE

∣∣∣1

= PE∣∣∣2+KE

∣∣∣2

SOLVE:12

(9000 kg) (0.3 m/s)2 =12k(0.05 m)2

k = 3.24×105 N/m

If this spring was used on all four corners of a car we’d have a combined spring stiffness of 4k or1.296×106 N/m Equating the force developed in the spring to counteract the force due to gravitygives us

mg = kx

(1400 kg)(9.81 m/s2) = (1.296×106 N/m)x

x = 1.06×10−2 m

164

4.2.42GOAL: Determine the maximum spring compression after a mass/spring combination strikes awall.GIVEN: v = 40 mph, weight = 3000 lb, k = 80, 000 lb/ftDRAW:

FORMULATE EQUATIONS:STATE 1:Just before collision, x = 40mph, spring uncompressed.STATE 2:At maximum spring compression (∆x), x = 0.

PE∣∣∣1+KE

∣∣∣1

= PE∣∣∣2+KE

∣∣∣2

SOLVE:

12

(3000 lb

32.2 ft/s2

)(40 mph

(88 ft/s60 mph

))2

=12(80, 000 lb/ft)∆x2

∆x = 2.0 ft

165

4.2.43GOAL: Determine if a falling spring restrained person will hit the ground.GIVEN: Spring constant, mass, initial height and dimensionsDRAW:

FORMULATE EQUATIONS:We’ll apply conservation of energy:

KE∣∣∣1+ PE

∣∣∣g1

+ PE∣∣∣sp

1

= KE∣∣∣2+ PE

∣∣∣g2

+ PE∣∣∣sp

2

SOLVE:Assume that initially her speed is zero.

KE∣∣∣1

= 0

PE∣∣∣g1

= mgh1

= (57 kg)(9.81 m/s2)(50m) = 2.80× 104 N·m

PE∣∣∣sp

1

= 0

Solve for her speed after falling 50 m

KE∣∣∣2

=12

mv22

= (28.5 kg)v22

PE∣∣∣g2

= 0

PE∣∣∣sp

2

=12

kx2 =12

(63 N/m)(50m− 20 m)2 = 2.84× 104 N·m

2.80× 104N·m = (28.5 kg)v22

+ 2.84× 104N·m

v22

= −14(m/s)2

A negative value for v22

implies an imaginary solution for v2. The conclusion is that she doesn’t

contact the ground, but rather comes to rest somewhat above the ground.

166

4.2.44GOAL: Find Favg during impact and formulate an expression for the minimum spring lengthGIVEN: Physical dimensions of the systemDRAW:

FORMULATE EQUATIONS:Non-conservative work is done during the drop and thus we’ll need to account for this in our energybalance:

KE1

+ PEg1

+ PESP1

− F∆x = KE2

+ PEg2

+ PESP2

(1)

SOLVE:a) Go from state 1 to just before impact (state 2, B moving at v

B)

12m

Bv2o + m

Bgh− 2F

f

[h− d−

(L−

mA

g

k

)]=

12m

Bv2B

+ mB

g

(L−

mA

g

k+ d

)

v2B

=2

mB

[12m

Bv2o + m

Bgh− 2F

f

[h− d−

(L−

mA

g

k

)]−m

Bg

(L−

mA

g

k+ d

)](2)

After impact we have

mB

vB

+ mA

vA

=(m

A+ m

B

)v

Using vA

= 0 gives

v =

(m

B

mA

+ mB

)vB

(3)

Impulse implies:

mB

vB− F∆t = m

Bv ⇒ F =

mB

vB−m

Bv

∆t

Where vB

and v are found from (2) and (3)b) Now we’ll go from just after impact, with A and B moving with speed v, to v

A= v

B= 0

12

(m

A+ m

B

)v2+m

Ag

(L−

mA

g

k

)+m

Bg

(L−

mA

g

k+ d

)+

12k

(m

Ag

k

)2

−2Ff

[L−

mA

g

k− L

min

]167

= mA

gLmin

+ mB

g(L

min+ d

)+

12k(L− L

min

)2

This can then be solved for Lmin

168

4.2.45GOAL: Find the average force acting between two mass particles during a collision and expressionsfor the distance traveled by the post-collision pair under different frictional conditions.GIVEN: System configuration.DRAW:

FORMULATE EQUATIONS:We’ll apply conservation of energy

KE∣∣∣1+ PE

∣∣∣1

= KE∣∣∣2+ PE

∣∣∣2

to determine the collision speed vc , conservation of momentum to calculate the post-collision speedvpc and work/energy

KE∣∣∣1+ PE

∣∣∣1+ W

1−2= KE

∣∣∣2+ PE

∣∣∣2

to calculate the distance traveled by the mass-pair.SOLVE:The spring is stretched to the same degree when m

1is released and when it contacts m

2. Thus it

needn’t be included in our energy balance. Going from release to just before collision we have12

m1v20

+ m1gr =

12

m1v2c

vc =√

v20

+ 2gr

Conservation of linear momentum then yields

m1vc = (m

1+ m

2)vpc

vpc =

(m

1

m1

+ m2

)vc

We know that the applied linear impulse (⇀F∆t) acts to speed up m

2and slow down m

1. We can

examine either mass to find |F |. Let’s choose m2. In the ⇀ı direction we have

F∆t = m2

(vpc − 0

)=

m1m

2vc

m1

+ m2

Hence we have|F | = m

1m

2vc(

m1+m

2

)∆t

Next we need to examine how far the two masses slide. Because the resisting force is a constant,the work done is simply the force multiplied by the displacement. Note that we’ll define x = 0 tobe the initial position of m

2and x is positive to the right (⇀ı direction).

12 k(r − L)2 + 1

2 (m1

+ m2)v2

pc − Fc∆x = 12 k[√

r2 + (∆x)2 − L]2

169

This is a single equation with the single unknown ∆x.Now consider a more complicated resisting force, that of Coulomb friction (F = µ

dN).

The friction force µdN varies with N and N varies with the position of the two masses:

N − (m1

+ m2)g + T sin θ = 0 ⇒ N = (m

1+ m

2)g − T sin θ

The spring force T is given by

T = k

[√r2 + (∆x)2 − L

]and sin θ is given by r√

r2+x2. Hence we have

N = (m1

+ m2)g − kr +

kLr√x2 + r2

The force due to friction is µdN and so our energy equation becomes

12 k(r − L)2 + 1

2 (m1

+ m2)v2

pc −∫∆x0 µ

dN dx = 1

2 k[√

r2 + (∆x)2 − L]2

This is again one equation in one unknown and can therefore be solved numerically for ∆x.

170

4.2.46GOAL: Find the impact speed of the pile driver with the ground.GIVEN: Mass of pile driver and dimensions, mass and orientation of support structure.DRAW:

FORMULATE EQUATIONS:We can use conservation of energy:STATE 1:

KE1

= 0 PE1

= mgL1cos 30◦

STATE 2:KE∣∣∣2

=12mv2 PE

∣∣∣2

= 0

SOLVE:

mgL1cos 30◦ =

12mv2

v =√

2gL1cos 30◦ =

√2(9.81 m/s2)(2m)(

√3

2) = 5.83 m/s

171

4.2.47GOAL: Find the speed of block A when it loses contact with the horizontal surface its slidingupon.GIVEN: System configuration, masses and spring/force characteristic.DRAW:


FORMULATE EQUATIONS: We’ll apply energy conservation. First, we need to deter-mine how to account for the spring’s potential energy. Comparing the force characteristic F =5mgL

0

(L− L

0

)with the characteristic of a linear spring (F = kx) lets us identify the spring’s

potential energy as

PE∣∣∣sp

=12

5mg

L0

(L− L

0

)2

where the linear spring’s potential energy is given by 12 kx2. The two blocks are connected by an

inextensible spring and therefore we know that they both have the same speed.Assuming that block B falls a distance ∆x when contact is lost, we start with a potential energyof mg∆x, zero kinetic and zero potential energy. When block B has fallen ∆x, both blocks aremoving with speed v and the spring is stretched to a length L from its initial length L

0:

mg∆x =12

mv2 +12

mv2 +12

5mg

L0

(L− L0)2 (1)

Our force condition for a loss of contact is given by

5mg

L0

(L− L0) cos θ = mg (2)

SOLVE:

(2)⇒ L =5L

0

4(3)

(1) ⇒ mv2 +12

5mg

L0

(L− L0)2 = mg∆x (4)

(3)⇒ ∆x =√

L2 − L20

=3L

0

4(5)

172

(5)→(4)⇒ mv2 +12

5mg

L0

L20

16= mg

3L0

4(6)

(6)⇒ v =√

1932gL

0

173

4.3 Power and Efficiency

174

4.3.1GOAL: Find the power output of a person climbing stairs.GIVEN: Rate of climb, height of a stair, number stairs per flight and flights per floor.DRAW:

FORMULATE EQUATIONS:For a constant power applied over a time t we have the relationship

P = W/t

where W is the work done.SOLVE:The force being applied by the climber is simply N = mg because the climbing is being done at aconstant speed.The change in height h over a single floor is given by

h = (7 in/step)(13 steps/flight)(2 flights/floor) = 182 in = 15.16 ft

P =W

t=

(168 lb)(15.16 ft)12 s

P = 212 lb· ft/s = 0.39 hp

175

4.3.2GOAL: Find the cost of raising an elevator and one passenger up 4 stories and the average powerrequired.GIVEN: Elevator efficiency, cost of electricity, mass of elevator/passenger, distance traveled andtime needed for the trip.DRAW:

FORMULATE EQUATIONS:For a constant power applied over a time t we have the relationship

P = W/t

where W is the work done and for an efficiency η we have

Pout = ηPin

, Wout = ηWin

SOLVE:The work done is equal to the force times the distance moved:

W = (270 kg + 70 kg)(9.81 m/s2)(19m) = 6.34×104 N·m = 6.34×10 kJ

Given an efficiency of 0.84 this means that the input work must have been

Win

=6.34×10 kJ

0.84= 7.54×10 kJ

The average power used is therefore

P =W

t=

7.54×10 kJ20 s

= 3.77 kW

The cost of electricity used is given by

cost = ($0.12/ kW· hr)(1 hr/3600 s)(20 s)(3.77 kW) = $0.0025

cost = 0.25 cents

176

4.3.3GOAL: Find cost to fill a water tank.GIVEN: Height of tank, quantity of water, efficiency of pump.DRAW:


W1−2

=mgh

(0.5)=

(1000 kg)(9.81 m/s2)(30m)(0.5)

= 5.89× 105 N·m

SOLVE:

1 kW · hr = (1000 N ·m/s)(3600 s) = 3.6× 106 N·m

# kW · hr needed =5.89× 105 N·m3.6× 106 N·m

= 0.16

Cost = (0.16)($0.15) = 2.5 cents

177

4.3.4GOAL: Find the power dissipated due to braking.GIVEN: Slope, weight of car, speed and acceleration.DRAW:


θ = tan−1 0.1 = 5.71◦


30 mph = 44 ft/s

We’ll use our power/speed relationship to determine the power: P = F s

Force balance: ms⇀b1 = −mg⇀ + F

⇀b1 + N

⇀b2

⇀b1 : ms = −mg sin θ + F (1)

⇀b2 : 0 = −mg cos θ + N (2)

SOLVE: The power dissipated by the brakes is equal to the force exerted on the car, F , times thecar’s speed, s. With s equal to zero we have

0 = −mg sin θ + F ⇒ F = mg sin θ

P = Fv = (3800 lb) [sin (5.71◦)] (44 ft/s) = 1.66 x 104 lb·ft/s = 30 hp

178

4.3.5GOAL: Find the power dissipated due to brakingGIVEN: Slope, weight of car, speed and accelerationDRAW:

⇀ı ⇀⇀b1 cos θ − sin θ⇀b2 sin θ cos θ

ASSUME: Motion remains along the hill’s surfaceFORMULATE EQUATIONS:

P = Fv

Force balance: ms⇀b1 = −mg⇀ − F

⇀b1 + N

⇀b2

⇀b1 : ms = mg sin θ − F (1)

⇀b2 : 0 = −mg cos θ + N (2)

θ = tan−1 (0.08) = 4.57◦

SOLVE: The power dissipated by the brakes is equal to the force exerted on the car, F , times thecar’s speed, s. Initially s = 0 and so F = mg sin θ. Call this F

1. When the car begins to decelerate

at 0.8g we have

m (−0.8g) = mg sin θ − F ⇒ F = mg (sin θ + 0.8)

Denote this value of F as F2.

Initial Power = F1v = (3200 lb) (sin (4.57◦)) (44 ft/s) = 1.12 x 104 lb·ft/s

Final Power = F2v = (3200 lb) (sin (4.57◦) + 0.8) (44 ft/s) = 1.24 x 105 lb·ft/s

Change in power is1.24 x 105 − 1.12 x 104

1.12 x 104⇒ 1000% change

179

4.3.6GOAL: Find how far up a mountain a single container of yoghurt will get you.GIVEN: Body’s effiiciency, mass of the body/bicycle and calories in yoghurt.DRAW:

FORMULATE EQUATIONS:For an efficiency η we have

Wout = ηWin

SOLVE:The work done is equal to the force times the distance moved. We neglect all external drag sourcesand simply consider the work done to elevate the body. Note that 1 “food” calorie is actually equalto 1000 calories (the calories for which 1 calorie=4.184 J).We’ll start with the energy supplied by the yoghurt:

E = 150 food cal = (150 food cal)4, 184 J

1 food cal= 627.6 kJ

We’re given that the body is 25% efficient and thus we will only extract (0.25)(627.6 kJ) = 156.9 kJfrom it.Equating the work done with the energy available yields

Fh = (75 kg)(9.81 m/s2)h = 156.9 kJ

h = 213m

180

4.3.7GOAL: Find the power output needed to go up a specified height in a specified time.GIVEN: Cyclist’s weight, height attained and travel time.DRAW:

FORMULATE EQUATIONS:We’ll be using our power/work relationship for constant work done:

P =W

t

SOLVE:

P =Fh

t=

(160 lb)(3500 ft)(44 min)(60 s/1 min)

= 212.1 ft· lb/s

P = (212.1 ft· lb/s) (1 hp)(550 ft· lb/s) = 0.386 hp

181

4.3.8GOAL: Find the average power put out in climbing a given hill at a certain rate and then determinethe change in time if an electric motor is added.GIVEN: Mass of the system, height of the mountain and initial time needed to reach the top.DRAW:

FORMULATE EQUATIONS:We’ll be using our power/work relationship for constant work done:

P =W

t

SOLVE:

P =Fh

t=

(175 lb)(3500 ft)(75 min)(60 s/1 min)

= 136.1 ft· lb/s

P = (136.1 ft· lb/s) (1 hp)(550 ft· lb/s) = 0.247 hp

Now we can modify our system by adding a 20 lb, 1/3 hp motor. This increases the overall hp to0.581 and the weight to 195 lb.

t =W

P=

(195 lb)(3500 ft)

(0.581 hp)550 ft· lb/s1 hp

t = 2137 s = 35.6 min

The addition of the electric motor would get me to the top 39.4 minutes faster.

182

4.3.9GOAL: Determine the coefficient a (part of a drag force formula). Also determine the change inspeed associated with a 10% increase in power.GIVEN: Form of the drag force, weight and speed of the cyclist and steady power output.DRAW:

FORMULATE EQUATIONS:Our diagram only shows the forces along the direction of travel, the traction force T and the dragforce F

d.

We’ll be using our power/speed relationship:

P = Fv

SOLVE:In a steady-state condition we have a balance between the drag force and the traction force:

T = Ff

= av2

Using this traction force in our power expression gives us

P = av3

We’re given that the power output is 0.35 hp and therefore can solve for a:

0.35 hp = 192.5 ft· lb/s = a(36.6 ft/s)3

a =192.5 ft· lb/s(36.6 ft/s)3

a = 3.90×10−3 lb· s2/ ft2

Increasing the power output by 10%, to 0.385 hp means we have

0.385 hp = 211.75 ft· lb/s = (3.90×10−3 lb· s2/ ft2)v3 ⇒ v = 37.9 ft/s = 25.8 mph

The change in the cyclist’s speed is just 0.8 mph , a bit over a 3 percent increase in speed for a 10percent increase in power.

183

4.3.10GOAL: Find the percentage increase in power output for double the speed.GIVEN: Relation between aerodynamic drag and speed.DRAW

ASSUME: Let’s assume that, for the two cases in which we’re interested, the force exerted by thebody exactly matches the drag force, such that it is not accelerating.FORMULATE EQUATIONS:

Drag: Fd

= av2 (1)

Power: P = Fdv (2)

SOLVE: Let v1

be the initial velocity, and v2

be double the initial velocity: v2

= 2v1. The

expressions for power at these two states are

P1

= Fdv1

= av31

P2

= Fdv2

= a(2v1)3

The percentage increase is given by

% increase =P

2− P

1

P1

× 100 =a(8v3

1− v3

1)

av31

× 100 =7av3

1

av31

× 100 = 700%

% increase = 700%

184

4.3.11GOAL: Find the power output needed to drive up a grade at a constant speed.GIVEN: m = 1200 kg, 5% grade, v = 20m/sDRAW:


θ = tan−1(0.05) = 2.86◦

ASSUME: Neglect air and road dragFORMULATE EQUATIONS:

Power P =⇀F · ⇀v

FBD=IRD N⇀b2 + F

⇀b1 −mg⇀ = mx = 0

SOLVE:⇀b1 : F −mg sin θ = 0 ⇒ F = mg sin θ

F = mg sin θ = mgθ = mg(0.0499)

P =⇀F · ⇀v = mg(0.0499)

⇀b1 · v

⇀b1

P = (1200 kg)(9.81 m/s2)(0.0499)(20m/s) = 11.7 kW

P = 11.7 kW = 15.8 hp

185

4.3.12GOAL: Find new velocity when grade increases.GIVEN: Constant power condition and grade change. Initial speed is 10 mph.DRAW:

FORMULATE EQUATIONS: We’ll apply work energy:

KE∣∣∣2+ PE

∣∣∣2

= KE∣∣∣1+ PE

∣∣∣1+ Wnc1−2 (1)

Definition of power: P =dW

dt(2)

The two angles of ascent are given by

θ1

= arctan 0.05 = 2.86◦, θ2

= arctan 0.06 = 3.43◦

Let the speed on the initial slope be given by v1

and on the greater slope be given by v2.

SOLVE:

(1) ⇒ 0 + mgh = 0 + 0 + Wcyclist

(3)

(2), (3) ⇒ P =dW

cyclist

dt= mg

dh

dt(4)(

dh

dt

)1

= v1sin θ1,

(dh

dt

)2

= v2sin θ2 (5)

(2), (4), (5) ⇒ P1

= P2⇒ v

2= v

1

sin θ1

sin θ2

(6)

v2

= 8.34 mph

186

4.3.13GOAL:Find the maximum attainable speed of the car and the horsepower at that speed.GIVEN: The car weighs 3500 lb, cr = 0.02, cr = 0.011 slug/ft, and F = 350 lbFORMULATE EQUATIONS:At the maximum speed x = 0.Thus a force balance gives us

0 = −crmg − k(x)2 + F

0 = −(0.02)(3500 lb)− (0.11 slug/ft)(x)2 + 350 lb

x = 160 ft/s = 109 mph

At this speed, the power is given by:

P = Fv = (350 lb)(160 ft/s) = 56, 000 lb·ft/s = 102 hp

187

4.3.14GOAL: Find the average horsepower generated by the cars.GIVEN: The cars are initially at rest and accelerate to 15 mph. µ

d= 0.55, µs = 0.8, and both

cars weigh 1500 lb.ASSUME: 35% of the weight is supported by the rear wheels and 65% is supported by the frontwheels.FORMULATE EQUATIONS:

mA

= mB

=3000lb

32.2 ft/s2= 93.2 slg, 15 mph = 22 ft/s

Car A: FdA

= 0.8(3000 lb)(0.65) = 1560 lb (drive force)

Car B: FdB

= 0.55(3000 lb)(0.65) = 1073 lb (drive force)

Apply force balance Car A:

(93.2 slg)xA

= 1560 lb

xA

= 16.7 ft/s2

To reach 15 mph (22 ft/s) requires:

(16.7 ft/s2)(tA

) = 22 ft/s

tA

= 1.31 s

hpA

= Fv = (1560 lb)(22 ft/s) = 34, 320 lb/ft = 62.4hp

hpavg = 62.4 hp1.31 s = 47.5 hp

Apply a force balance to Car B:

(93.2 slug)xB

= 1073 lb

xB

= 11.5 ft/s2

To reach 15 mph (22 ft/s) requires:

(11.5 ft/s2)(tB

) = 22 ft/s

tB

= 1.91 s

hpB

= Fv = (1073 lb)(11.5 ft/s) = 12340 lb/ft = 22.4 hp

hpavg = 22.4 hp1.91 s = 12 hp

Car B takes more than 50% as long as car A to reach 15 mph. It seems that slipping is a far lesseffective way of transferring power from the engine to the car than no-slip. The average horsepowerfor the slip case is lower both because the force is lower (1073 lb vs 1560 lb) and the time overwhich the average was taken is longer.

188

4.3.15GOAL:What is the horsepower developed after 3 seconds?What is the average horsepower over the distance traveled?DRAW:


Force balance, ⇀ı : mx = F (1)

Power: P = Fx (2)

SOLVE:If the vehicle is rolling without slipping then the force sustainable by the frictional interface is

Fdrive

= mgµs = (1500 kg)(9.81 m/s2)(0.9) = 13, 244 N

This implies an acceleration of

(1500 kg)x = Fdrive

= 13, 244 N

x = 8.83 m/s2

After 3 seconds the vehicle is traveling at

x = (8.83 m/s2)(3 s) = 26.5 m/s

The power is equal to the force times the velocity:

P = Fx = (13244 N)(26.5 m/s) = 351 kW

In terms of horsepower this is

P = (351× 103 W)(

1 hp746 W

)= 470 hp

P = 470 hpx increases linearly with time. The average hp is therefore

hpavg = 4702 = 235 hp

189

4.3.16GOAL: Determine a motor’s effiiciency.GIVEN: Mass of load, speed of load, slope’s inclination and electrical power input.DRAW:

FORMULATE EQUATIONS:We’ll use our FBD to determine the tension applied to the load and then use P = Tv to the findthe power and η = Pout/P

into find the efficiency.

SOLVE:From the FBD we have

T −mg sin θ = 0

T = mg sin θ = (1000 kg)(9.81 m/s2)(0.5) = 4905 N

P = Tv = (4905 N)(2 m/s) = 9810N·m/s = 9810 W

η =9810 W

18×103 W= 0.545

Thus we see that η = 0.545 and we have an efficiency of 54.5 percent .

190

4.3.17GOAL: Find the amount of power and work required to move gravel.GIVEN: Hopper moves at 1.5 m/s.DRAW:

FORMULATE EQUATIONS:We’ll use the power/speed formulation

P = Fv

SOLVE:

θ = tan−1 8 ft22 ft

= 16.5◦

P = Fv = (100 lb + 30 lb)(1.5 ft/s) cos(16.5◦) = 187 lb·ft/s

P = 187 lb·ft/s = 0.34 hp

W = F∆x = (130 lb)(27 ft) = 3510 lb·ft

191

4.3.18GOAL: Find the input power to ride at steady speed and the effect of road grade on speed.GIVEN: Drag force and cyclist’s efficiency(a)DRAW:

FORMULATE EQUATIONS: Newton’s law decomposed in the ⇀ı and ⇀ directions

FT − a1 − a2v2 = mx (1)

N −mg = 0 (2)

SOLVE: For a constant speed x = 0. Substituting in (1) gives us

FT = a1 + a2v2

The normal force N does no work so we need only consider FT . Velocity is given by v = 25 mph =11.2 m/s. The power output is

Pout = FT v = (3.3 N + (0.24 N· s2 /m2)(11.2)2)(11.2 m/s) = 372W (3)

The efficiency is 96% so the power input is

Pin =372 W0.96

= 387W = 0.52 hp (4)

(b)DRAW:

FORMULATE EQUATIONS: A force balance decomposed in the⇀b1 and

⇀b2 directions gives

us

FT −mg sin θ − a1 − a2v2 = 0 (5)

N −mg = 0 (6)

We’re looking for a steady speed and thus the acceleration is zero.SOLVE: Use (5) to find

FT = mg sin θ + a1 + a2v2 = (85 kg)(9.81 m/s2)(0.05) + 3.3 N + (0.24 N· s2 /m2)v2 = 44.9 N + 0.24v2

(7)

192

From part (a) we know the power driving the bike/person up the hill is 372 W.

372 W = (44.9 N + (0.24 N· s2 /m2)v2)v (8)

v = 6.68 m/s = 14.9 mph

This is a huge decrease in speed and highlights the well known fact that when you ride up a hillyou go a lot slower that when you’re riding on the flats.

193

4.3.19GOAL: Find average power output.GIVEN: Average speed is 11 mph, the road has an average grade of 5.5%, m = 76 kg, road dragforce is 2.5 NDRAW:


Force balance,⇀b1: F

traction− F

drag−mg sin θ = 0

sin θ = sin(tan−1(0.055)) = 0.055 rad

11 mph = 17.7 kph = 4.92 m/s

Ftraction

= mg sin θ + 2.5 N + (0.2 N·s2/m2)v2

Ftraction

= (9.81 m/s)(76 kg)(0.055 rad)+2.5 N+(0.2 N·s2/m2)(4.92 m/s)2

Power = Ftraction

v = (48.3 N)(4.92 m/s) = 237W = 0.32 hp

194

4.3.20GOAL:DRAW:


Force balance,⇀b1: F − F

drag−mg sin θ = 0

F = mg sin θ + 2.5 N + (0.21 N· s2 /m2)v2

F = (9.81 m/s2)(76 kg)(0.06) + 2.5 N + (0.21 N· s2 /m2)v2

F = 47.2 N + (0.21 N· s2 /m2)v2

P = Fv = (47.2 N + (0.21 N· s2 /m2)v2)v

We can convert our power output from hp to watts:

0.31 hp = 0.31 hp(746W/hp) = 231 W

231 W = (47.2 N)v + (0.21 N· s2 /m2)v3

(0.21 N· s2 /m2)v3 + (47.2 N)v − 231 W = 0

v = 4.50 m/s

Now consider the case of no road drag at all.

F = mg sin θ + 0.21v2 = (0.21 N· s2 /m2)v2 + 44.7 N

231 W = [44.7 N + (0.21 N· s2 /m2)v2]v

(0.21 N· s2 /m2)v3 + (44.7 N)v − 231 W = 0

v = 4.69 m/s

Eliminating all road drag increases my velocity by 4.3%

4.3% increase

195

4.3.21GOAL: Determine an electric motor’s efficiency.GIVEN: Mass of load, distance traveled and time needed to lift load and electrical power suppliedto the motor.DRAW:

FORMULATE EQUATIONS:We’ll determine the work done (force applied over distance), then use P = W/t to find the averagepower and finally use η = Pout/P

into determine the efficiency.

SOLVE:The work done is given by

W = Th = (200 kg)(9.81 m/s2)(4m) = 7848 J

P =W

t=

7848 J3 s

= 2616W

η =2616 W3000 W

= 0.872

Thus we see that η = 0.872 and we have an efficiency of 87.2 percent .

196

4.3.22GOAL: Plot the power acting on a block.GIVEN: Mass of block, velocity profile and inclination of slope.DRAW:

FORMULATE EQUATIONS:We’re asked to find the overall power applied to the block. Two forces are acting - gravity and thetension in the rope. Because we’re given the acceleration of the block, we don’t actually need todistinguish between the two. We know from a force balance that

F = ms

where F represents the sum of the forces acting in the direction of motion. Using this force inP = Fv will allow us to determine the power.SOLVE:

Force balance: F = ms = (125 kg)(

23

m/s2)

= 83.3 N

The power acting on the block is given by

P = (83.3 N)s

We can see that the power is a linearly increasing function of s.The acceleration of the block is constant and thus we have

s = st =(

23

m/s2)

t

and

P = (83.3 N)(

23

m/s2)

t = (55.5 W/s)t

197

198

4.3.23GOAL: Determine the zero to 30 mph time for a car that has no drag forces and puts out itsmaximum power at all times.GIVEN: Mass of the car and the maximum horsepower of its engine.DRAW:

FORMULATE EQUATIONS:To determine the time needed we’ll use our speed/time/constant power relationship

∆t =mv2

2P

where v is the final speed (car starts from zero) and P is the applied power.SOLVE:

m =3200 lb

32.2 ft/s2= 99.4 slg

P = 200 hp = (200 hp)550 ft· lb/s

1 hp= 110, 000 ft· lb/s

∆t =mv2

2P=

(99.4 slg)(44 ft/s2)2

2(110, 000 ft· lb/s= 0.875 s

∆t = 0.875 s

199

4.3.24GOAL: Plot the rate at which energy is removed from a body that’s slowing at a specified rate.GIVEN: Mass of the body and time to slow to zero.DRAW:

FORMULATE EQUATIONS:We’ll use the power/time relationship

P = Fbv

SOLVE:The body slows from 100 km/hr to zero in 4 s, which implies a constant acceleration of

a =0− 27.7 m/s

4 s= −6.94 m/s2

Fb

= ma = (1600 kg)(−6.94 m/s2) = −11, 104 N

P = Fbv = (−11, 104 N)v

200

4.3.25GOAL: Determine the drag coefficient c

dacting on a car.

GIVEN: Relevant efficiencies, fuel usage and energy density of gasoline.DRAW:

FORMULATE EQUATIONS:What we’ll do is compute the total power in a gallon of gasoline. We’ll then reduce this by thecombined drivetrain and combustion efficiency (η

dtand ηcom , respectively) to get the total work

supplied to move the car. We’ll next take our expression for the force acting on the car (F ) thatcounteracts the drag force. Multiplying by the distance traveled will give us the work. Equatingthese two energies will allow us to compute c

d.

SOLVE:The total energy in a gallon of gasoline is 1.25×108 J and the delivered energy is

Edel

= (1.25×108 J)ηdt

ηcom = (1.25×108 J)(0.8)(0.28) = 2.8×107 J

The car burns 2.7 gal/hr and thus delivers a total energy of

E = 2.7(2.8×107 J) = 7.56×107 J

The energy used by the car is given by the force times the distance traveled. In one hour we’ll use2.7 gal of gasoline and travel 60 miles = 9.66×104 m. The work done is thus

W = Fd = 0.5ρAcdv2d = 0.5(1.2 kg/m3)(2m2)c

d(26.8 m/s)2(9.66×104 m)

W = cd(8.34×107 kg·m2 /s2)

Equating the work done and the energy used gives us

7.56×107 J = cd(8.34×107 kg·m2 /s2)

cd

= 0.91

201

4.3.26GOAL: Determine power supplied to an electric motor.GIVEN: Mass of load, distance traveled and time needed to lift load and motor’s efficiency.DRAW:

FORMULATE EQUATIONS:We’ll determine the work done (force applied over distance), then use P = W/t to find the averagepower and finally use η = Pout/P

into determine the input power given the known efficiency.

SOLVE:Two ropes support the mass, giving us an equilibrium (zero acceleration) condition of

2T = mg ⇒ T =mg

2We know from our pulley kinematics that the reel will pull rope into it at twice the speed withwhich the load is lifted. Thus, if the load moves up 3 m in 5 s, its speed is 0.6 m/s and the ropefeeds into the reel at 1.2 m/s.The power done at the reel is given by

P = Tv =(50 kg)(9.81 m/s2)

2(1.2 m/s) = 2.94×102 W

η = 0.60 =2.94×102 W

Pin

⇒ Pin

= 4.91×102 W

Pin

= 4.91×102 W

202

4.3.27GOAL:Determine the maximum acceleration you can achieve on your bicycle.Determine the time needed to catch up to other person and distance traveled.GIVEN: Mass of you and the bicycle, the drag forces acting and your maximum power output.FORMULATE EQUATIONS:

Force balance: (F − Fdrag

)⇀ı + (N −mg)⇀ = mx ⇀ı

⇀ı : F − Fdrag

= mx ⇒ x =1m

(F − Fdrag

) (1)

(a):SOLVE:In steady state conditions we have:

F = Fdrag

= 2.8 N + (0.22 N·s2/ m2)v2 (2)

v = 15mph = 24.2 kph = 6.7 m/s (3)

(2),(3)⇒ F = 2.8 N + (0.22 N·s2/m2)(6.7 m/s)2 = 12.7 N (4)

P = Fv = (12.7)(6.7) = 85 W

Steady state power is 85 W

Maximum power is 0.4 hp = (0.4 hp)(745.7 W/hp) = 298W.At 6.7 m/s this implies a force of

F (6.7 m/s) = 298W ⇒ 44.5 N (5)

(5)→(1)⇒ x =(

182kg

)(44.5 N− 12.7 N) = 0.388 m/s2

(b):To determine the positions and speed versus time we need the governing equation for the bicycle.The maximum force the cyclist can produce is given by

Fmax x = 298W ⇒ F =298 W

x

Our equation of motion is therefore given by

x =1

82 kg(Fmax − F

drag) =

182 kg

(298 W

x− 2.8 N− (0.22 N·s2/m2)x2

)The following shows a simple MATLAB file that contains this equation as well as plots of x and xversus time. The time needed to reach 20 mph is 8.86 s.

203

204

4.3.28GOAL: Plot the rate at which energy is removed from a body that’s slowing at a specified rate.GIVEN: Mass of the body and distance needed for the deceleration.DRAW:

FORMULATE EQUATIONS:We’ll use the power/speed relationship

P = Fbv

and our formula that relates distance traveled, speed and acceleration (under a constant accelerationassumption):

a(x2− x

1) =

12

(v22− v2

1

)(1)

SOLVE:The body slows from 60 mph (88 ft/s) to zero in 120 ft with a constant deceleration.

m =

(3500 lb

32.2 ft/s2

)= 108.7 slg

(1) ⇒ a(120 ft) =12

(0− (88 ft/s)2

)a = −32.26 ft/s2

The force acting on the body (Fb) times the speed of the body gives us the power and the force

acting on the body is simply the mass times the acceleration:

P = Fbv = mav = (108.7 slg)(−32.26 ft/s2)v = −(3507 slg· ft/s2)v

The easiest way to plot this versus displacement, as asked for, is to use our velocity/displacementrelationships

v(t) = 88 ft/s− (32.26)t, x(t) = (88 ft/s)t− (32.26 ft/s2)t2

2,

compute v and x from t = 0 to t = 2.727 s (the start and finish times of the braking maneuver)and then use v to compute P and plot versus x. Doing so in MATLAB leads to the following plotof P vs x

205

206

4.3.29GOAL: Determine the power delivered to a spinning wheel.GIVEN: System configuration, relevant masses, angular velocity of the wheel and dynamics coef-ficient of friction.DRAW:

FORMULATE EQUATIONS:We’ll use the power/speed relationship

P = Fv

where F is the force due to friction and v is the speed of the outer periphery of the wheel.SOLVE:The mass is slipping against the periphery of the wheel and therefore feels a frictional forceF = µN = µmg cos θ (pointing in a clockwise direction) and a gravitational force mg sin θ point-ing counter-clockwise. These forces have to balance so that the mass stays fixed in a particularorientation:

µmg cos θ = mg sin θ

tan θ = µ ⇒ θ = 11.3◦

We now know the steady-state position of the mass with respect to the wheel.As the FBD shows, the wheel has to supply power to counter the force that’s continually trying torotate the wheel counter-clockwise. The power required is equal to this frictional force F multipliedby the slip speed:

P = (µmg cos θ)vslip

= 0.2(0.08 kg)(9.81 m/s2)(0.981)(15 rad/s)(0.1 m)

P = 0.231 W

207

Documents

ch04