Module 7Work and KE

2005 Pearson Education South Asia Pte Ltd

MECHANICS

FE1001 Physics I NTU - College of Engineering

1. Units, Physical Quantities and Vectors

2. Motion Along A Straight Line3. Motion in 2 or 3 Dimensions4. Newton’s Law of Motion

5. Applying Newton’s Laws6. Work and Kinetic Energy7. Potential Energy and Energy Conservation


MECHANICS

FE1001 Physics I NTU - College of Engineering

8. Momentum, Impulse, and Collisions

9. Rotation of Rigid Bodies10. Dynamics of Rotational

Motion11. Equilibrium and Elasticity12. Gravitation13. Periodic Motion14. Fluid Mechanics

6. Work and Kinetic Energy


Chapter Objectives

• Understanding the concept of work in daily lives.• The difference between work and kinetic energy.• The relation of work and energy with varying forces.• Understanding of power.



Chapter Outline

1. Work


3. Work and Energy with Varying Forces

4. Power



6.1 Work

• Work is any activity that requires muscular or mental effort.• In physics, the total work done on a particle by all forces

that acts on it equals the change of kinetic energy.• The relationship holds true even when the forces acting on

the body is not constant.• When a body moves, a constant force acts on it in the

same direction as the displacement .

Fur

sr



6.1 Work

• We define the work W done by this constant force under these circumstances as the product of the force F and the displacement s.

(6.1)W Fs=

(constant force in direction of straight-line displacement)



6.1 Work

• Don’t confuse W (work) with w (weight) as they have different quantities.

• SI unit of work is in joule.

( )( )1 joule 1 newton 1 meter or 1 J = 1 N m= ⋅

• For British unit,

1 J = 0.7376 ft lb ⋅1 ft lb = 1.356 J⋅



6.1 Work

• When a person pushes a car at an angle as shown, the force and the displacement will have different directions.

• Thus we define the work as a product of the component and the magnitude of the displacement.

φFur

sr



6.1 Work

• When a constant force acts at an angle to the displacement , the work done by the force is

φFur

sr

cos (6.2)W Fs φ=

(constant force, straight line displacement)

• When Eq. (6.2) is in the form of scalar product of 2 vectors,

(6.3)W F s= ⋅uur ur r

(constant force, straight line displacement)



6.1 Work

• Work is a scalar quantity even though it’s calculated by using 2 vector quantities (force and displacement).



Example 6.1 Work done by a constant force

a) Steve exert a steady force of magnitude 210N (about 47lb) on the stalled car in figure shown as he pushed it a distance of 18m. The car also has a flat tire, so to make the car track straight Steve must push at an angle of 30o to the direction of motion. How much work does Steve do? b) In a helpful mood, Steve pushes a second stalled car with steady force The displacement of the car is How much work does Steve do in this case?

(160 ) (40 ) .F N i N j= −r

ˆ ˆ(14 ) (11 ) .s m i m j= +r



Example 6.1



Example 6.1 (SOLN)

Identify and Set upIn both part (a) and (b), the target variable is the work W done by Steve. In each case the force is constant and the displacement is along a straight line, so we can use Eq.(6.2) or (6.3). The angle between and is given explicitly in part (a), so we can apply Eq.(6.2) directly. In part (b) the angle isn’t given. So we’re better off calculating the scalar product in Eq.(6.3) form the components of and , as in Eq.(1.21).

Fr

sr

Fr

sr



Example 6.1 (SOLN)Executea)From Eq.(6.2),

The components of are Fs = 160N and Fy = -40N, and the components of are x = 14m and y = 11m. Hence, using Eqs.(1.21) and (6.3),

3cos (210 )(18 )cos30 3.3 10W Fs N m Jφ= = °= ×

Fr

sr

x yW F s F x F y= = +r rg

(160 )(14 ) ( 40 )(11 )N m N m= + −31.8 10 J= ×Evaluate

Our results show that 1 joule is a rather small amount of work.



6.1 Work

• A constant force can do positive, negative or zero work depending on then angle between and the displacement .

• When force has component in direction of displacement, work done is positive.

Fur

sr F

ur

0 90φ< < °



6.1 Work

• When force has component opposite to displacement, work done is negative.

90 180φ< < °



6.1 Work

• When force is perpendicular to displacement, work done is zero.

90φ = °



6.1 Work

• An object do no work if has no displacement even when a force is apply on it.

• Below is the illustration on positive, zero and negative work.



6.1 Work

• Negative work comes from Newton’s third law of motion. • When you catch a ball, your hand and ball move together with the same displacement, .• The work done by the ball on your hand is positive but the your hand exerts an equal and

opposite to the ball’s displacement, which is negative.

sr

H on B B on HF F=−ur ur



6.1 Work

• Always specify what force is doing the work you are talking about.• For example the work done by lifting a book is positive but the work done by

the gravitational force on a book is negative, as downward gravitational force is opposite to the upward displacement.



Example 6.2 Work done by several forces

Farmer Benton hitches his tractor to a sled loaded with firewood and pulls it a distance of 20m along level ground. The total weight of sled and load is 14,700N. The tractor exerts a constant 5000-N force at an angle of 36.9o above the horizontal, as shown in figure. There is a 3500-N friction force opposing the sled’s motion. Find the work done by each force acting on the sled and the total work done by all the forces.



Example 6.2 Work done by several forces



Example 6.2 (SOLN)

IdentifyAll of the forces are constant and the displacement is along a straight line, so we can calculate the work using the formulas given in this section. We’ll find the net work in two ways: (1) by adding together the work done on the sled by each force and (2) by finding the amount of work done by the net force on the sled. We’ll find the net force using the techniques described in Chapter 5.



Example 6.2 (SOLN)

Set upAs we are working with forces, our first steps are to draw a FBD showing all of the forces acting on the sled and to choose a coordinate system.



Example 6.2 (SOLN)

Set upFor each of the four forces – weight, normal force, force of the tractor, and friction force – we know the angle between the displacement (which is in the positive x-direction) and the force. Hence we can calculate the work done by each force using Eq.(6.2).

To find the net force, we’ll add the components of the four forces. Newton’s second law tells us that because the sled’s motion is purely horizontal, the net force will have only a horizontal component.



Example 6.2 (SOLN)

ExecuteThe work Ww done by the weight is zero because its direction is perpendicular to the displacement. (The angle between the force of gravity and the displacement is 90o, and the cosine of the angle is zero.) For the same reason, the work Wn done by the normal force is also zero. So Ww = Wn = 0. (Incidentally, the normal force is not equal in magnitude to the weight; see Example 5.16 of Section 5.3.)



The friction force is opposite to the displacement, so for this force and . The work Wf done by the friction force is

Example 6.2 (SOLN)ExecuteThat leaves the force Ft exerted by the tractor and the friction force f. From Eq.(6.2) the work WT done by the tractor is

cos (5000 )(20 )(0.800) 80,000 .T TW F s N m N mφ= = =80kJ=

fr

180φ = ° cos 1φ =−

cos180 (3500 )(20 )( 1) 70,000 .fW fs N m N m= °= − =−70kJ=−



Example 6.2 (SOLN)

Execute

The total work Wtot done on the sled by all forces is the algebraic sum of the work done by the individual forces:

0 0 80 ( 70 )tot w n T fW W W W W kJ kJ= + + + = + + + −

10kJ=



Example 6.2 (SOLN)

Execute

In the alternative approach, we first find the vector sum of all the forces (the net force) and then use it to compute the total work. The vector sum is best found by using components. From Fig.b,

cos ( ) (5000 )cos36.9 3500x TF F f N Nφ= + − = °−∑500N=

sin ( )y TF F n wφ= + + −∑(5000 )sin 36.9 14,700N n N= °+ −



Example 6.2 (SOLN)

Execute

We don’t really need the second equation; we know that the y-component of force is perpendicular to the displacement, so it does not work. Besides, there is no y-component of acceleration, so has to be zero anyway. The total work is therefore the work done by the total x-component:

yF∑

( ) ( ) (500 )(20 ) 10,000tot xW F s F s N m J= = = =∑ ∑r rg

10kJ=



Example 6.2 (SOLN)

Evaluate

We get the same result for Wtot with either method, as we should.

Note that the net force in the x-direction is not zero, and the sled must accelerate as it moves. In Section 6.2 we’ll return to this example and see how to use the concept of work to explore the sled’s motion.



Example 6.3 Total work when velocity is constant

A electron moves in a straight line toward the east with a constant speed of 8 x 107 m/s. it has electric, magnetic, and gravitational forces acting on it. Calculate the total work done on the electron during a 1-m displacement.



The electron moves with constant velocity, so its acceleration is zero. By Newton’s second law, the vector sum of the forces is also zero. Therefore the total work done by all the forces (equal to the work done by the vector sum of all the forces) must be zero. Individual forces may do nonzero work, but that’s not what the problems asks for.

Example 6.3 (SOLN)



6.2 Work and Kinetic Energy

• The total work done on a body by external forces is related to the body’s displacement, that is, no change in its position.

• But the total work is also related to the changes in the speed of the body.

• For example when a block is sliding on a frictionless table, the net force causes the speed to increase and does positive work.




• Again the net force causes the speed to increase and does positive work, when there is a change of direction of force.




• When the net force opposes the displacement and causes the speed to decrease, it does negative work.




• The net force is zero and does no work, thus the speed is zero.




• Consider that a mass moving along x-direction with a constant acceleration, this time speed changes from to .

• The mass undergoes a displacement of .• Using Eq. (2.13), 1v 2v

2 1s x x= −

2 22 1 2 xv v a s= +

2 22 1

2xv v

as−=




• When multiply by and equate to the net force,

m xma

2 22 1

2xv v

F ma ms−= =

2 22 1

1 1 (6.4)

2 2Fs mv mv= −

where = total work done by all the forces acting on the particle, = kinetic energy,

FstotW

212

mv K




• Therefore,

21 (6.5)

2K mv=

( definition of kinetic energy)

• From Eq. 6.4, it says that the work done by the net force on a particle equals the change in the particle’s kinetic energy:

2 1 (6.6)totW K K K= − =Δ

( work-energy theorem)



1. When , , the kinetic energy increases. The particle is going faster at the end of the displacement than at the beginning.

2. When , , the kinetic energy decreases and the speed is less after displacement.

3. When , , the speed remain unchanged.


• The work-energy theorem gives the following agreement:

0totW <

0totW > 2 1K K>

2 1K K<

0totW = 2 1K K=



6.2 Work and Kinetic Energy• Note that kinetic energy does not tell us about the direction of motion.• The SI unit of kinetic energy and work is

• The British unit is

2 2 21 J = 1 N m = 1 (kg m/s ) m = 1 kg m /s⋅ ⋅ ⋅ ⋅

2 2 21 ft lb = 1 ft slug ft/s = 1 slug ft /s⋅ ⋅ ⋅ ⋅




Problem-solving strategy (Work and Kinetic Energy)

• IDENTIFY

1.Use work-energy theorem for a body having moving to at a different point.

2.This theorem only used for problems which do not involve time.





• SET UP

1.Choose initial and final position and draw free-body diagram, showing all forces acting on the body.

2.Choose a coordinate system.

3.List all the known and unknown quantities, decide which unknowns are your target variables.





• EXECUTE

1.Calculate the work done by each force.

2.Use Eq. (6.2) and (6.3) if the force is constant and the displacement is straight line.

3.Check the sign of the work for each force.

4.Add the amounts of work done by each force to find the total work, . Be careful with signs! totW





• EXECUTE

5. Write expressions for the initial and final kinetic energies, and .

6. Use to solve for target variables.

1K 2K

2 1totW K K= −





• EVALUATE

1.Check whether your answer makes any physical sense.

2.Remember can never be negative.212

K mv=



Example 6.4 Using work and energy to calculate speed

Let’s look again at the sled and the numbers at the end of Example 6.2. Suppose the initial speed v1 is 2.0m/s. What is the speed of the sled after it moves 20m?



Example 6.4 (SOLN)

We’ll use the work-energy theorem, since we are given the initial speed v1 = 2.0m/s and want to find the final speed v2. Figure shows the FBD from Example 6.2. The motion is in the positive x-direction.

Identify and Set up



Example 6.4 (SOLN)

ExecuteWe already calculated the total work done by all the forces in Example 6.2, where we found Wtot = 10kJ. Hence the kinetic energy of the sled and its load must increase by 10kJ.

To write expressions for the initial and final kinetic energies, we need the mass of the sled and load. We are given that the weight is 14,700N, so the mass is

2/ (14,700 )(9.8 / ) 1500m w g N m s kg= = =



Example 6.4 (SOLN)

ExecuteThen the initial kinetic energy K1 is

2 2 2 21 1

1(1500 )(2.0 / ) 3000 /

2 2K mv kg m s kg m s= = = g

3000J=The final kinetic energy K2 is

Where v2 is the unknown speed we wan to find. Equation (6.6) gives

2 22 2 2

1 1(1500 )

2 2K mv kg v= =

2 1 3000 10,000 13,000totK K W J J J= + = + =



Example 6.4 (SOLN)

ExecuteSetting these two expressions for K2 equal, substituting , and solving for v2, we find

2 4.2 /v m s=

The total work is positive, so the kinetic energy increases ( K2 > K1 ) and the speed increases ( v2 > v1 ).This problem can also be done without the work-energy theorem. We can find the acceleration from and then use the equations of motion for constant acceleration to find v2.

2 21 1 /J kg m s= g

Evaluate

F ma=∑r r



Example 6.4 (SOLN)

Evaluate

( )x

5000 cos36.0 3500a = a

1500xF N N

m kg

° −= =∑

then

Since the acceleration is along the x-axis,

20.333 /m s=

2 2 2 22 1 2 (2.0 / ) 2(0.333 / )(20 )asv v m s m s m= + = +

2 217.3 /m s=

2 4.2 /v m s=



This is the same result that we obtained with the work-energy approach, but there we avoided the intermediate step on finding the acceleration. You will find several other examples in this chapter and the next that can be done without using energy considerations but that are easier when energy methods are used. When a problem can be done by two different methods, doing it by both methods (as we did in this example) is a very good way to check your work.

Example 6.4 (SOLN)

Evaluate



In a pile driver, a steel hammerhead with mass 200kg is lifted 3.00m above the top of a vertical I-beam being driven into the ground. The hammer is then dropped, driving the I-beam 7.4 cm farther into the ground. The vertical rails that guide the hammerhead exert a constant 60-N friction force on the hammerhead. Use the work-energy theorem to find a) the speed of the hammerhead just as it hits the I-beam and b) the average force the hammerhead exerts on the I-beam. Ignore the effects of the air.

Example 6.5 Force on a hammerhead



Example 6.5 Force on a hammerhead



This problem is an ideal candidate for the work-energy theorem, as it relates a body’s speed at different locations and the forces acting on the body. In this problem there are three positions of interest: point 1, where the hammerhead starts from the rest; point 2, where it first contacts the I-beam; and point 3, where the hammerhead comes to a halt. We have two unknowns – the hammerhead’s speed at point 2, and the force the I-beam exerts between points 2 and 3 – we’ll apply the work-energy theorem twice: once for the motion from 1 to 2 and once for the motion from 2 to 3.

Example 6.5 (SOLN)

Identify



Figure (b) is a FBD showing the vertical forces on the hammerhead as it falls from point 1 to point 2. (because the displacement is vertical, we ignore any horizontal forces that may be present because they do no work,) For this part of the motion, our target variable is the hammerhead’s speed v2.

Example 6.5 (SOLN)

Set up



The FBD in Fig. c shows the vertical forces on the hammerhead during the motion from point 2 to point 3. In addition to the forces shown in Fig. B, the I-beam exerts an upward normal force of magnitude n on the hammerhead. This force actually varies as the hammerhead comes to a halt, but for simplicity we’ll treat n as a constant. Hence n represents the average value of this upward force during the motion.

Example 6.5 (SOLN)

Set up



Our target variable for this part of the motion is the force that the hammerhead exerts on the I-beam: it is the reaction force to the normal force exerted by the I-beam, so by Newton’s third law its magnitude is also n.

Example 6.5 (SOLN)

Set up



Example 6.5 (SOLN)

Executea) From point 1 to point 2, the vertical forces are the downward weight w = mg = (200kg)(9.8m/s2) = 1960N and the upward friction force f = 60N. Thus the net downward force is w – f = 1900N. The displacement of the hammerhead from point 1 to point 2 is downward and equal to s12 = 3.00m. The total work done on the hammerhead as it moves from point 1 to point 2 is then

12( ) (1900 )(3.00 ) 5700totW w f s N m J= − = =



At point 1 the hammerhead is at rest, so its initial kinetic energy is K1 zero. Equation (6.6) gives

Example 6.5 (SOLN)

Execute

22 1 2

10

2totW K K mv= − = −

22 2(5700 )

7.55 /200

totW Jv m s

m kg= = =

This is the hammerhead’s speed at point 2, just as it hits the beam.



b) As the hammerhead moves downward between points 2 and 3, the net downward force acting on it is w – f – n (Fig. C). The total work done on the hammerhead during this displacement is

Example 6.5 (SOLN)

Execute

23( )totW w f n s= − −

The kinetic energy for this part of the motion is K2 where from part (a) equals 5700J (the total work done on the hammerhead as it moved from point 1 to point 2).



The final kinetic energy is K3 = 0, since the hammerhead ends at rest. Then, from the work-energy theorem,

Example 6.5 (SOLN)

Execute

23 3 2( )totW w f n s K K= − − = −

3 2

23

( )K Kn w f

s

−= − −

(0 5700 )1960 60

0.074

J JN N

m

−= − −

79,000N=



The downward force that the hammerhead exerts on the I-beam has this same magnitude, 79,000N (about 9 tons) – more than 40 times the weight of the hammerhead.

Example 6.5 (SOLN)

Execute



The total change in the hammerhead’s kinetic energy during the whole process is zero; a relative small force does positive work over a large distance, and then a much larger net force does negative work over a much smaller distance. The same thing happens if you speed up your car gradually and then drive it into a brick wall. The very large force needed to reduce the kinetic energy to zero over a short distance is what does the damage to your car – and possibly to you.

Example 6.5 (SOLN)

Evaluate




The Meaning of Kinetic Energy

• Example 6.5 gives the physical meaning of kinetic energy.

• To accelerate a mass m from rest (zero kinetic energy) up to speed v, the total work done on it must equal the change in kinetic energy from zero to ; 21

2K mv=

0totW K K= − =




The Meaning of Kinetic Energy

1. Kinetic energy of a particle is equal to the total work that was done to accelerate if from rest to its present speed.

2. Kinetic energy of a particle is equal to the total work that particle can do in the process of being brought to rest.

Kinetic Energy can be defined in 2 ways as below:



Example 6.6 (SOLN)



Example 6.6 (SOLN)

If you simply use the mathematical definition of the kinetic energy,

the answer to this problem isn’t immediately obvious. The iceboat of mass 2m has greater mass, so you might guess that the larger iceboat attains a greater kinetic energy at the finish line. But the smaller iceboat, of mass m, crosses the finish line with greater speed, and you might guess that this iceboat has the greater kinetic energy. How can we decide?

21

2K mv=



Example 6.6 (SOLN)

The correct way to approach this problem is to remember that the kinetic energy of a particle is just equal to the total work done to acceleration it from rest. Both iceboat travel the same distance s, and only the horizontal force F in the direction of motion does work on either iceboat. Hence the total work done between the starting line and the finish line is the same for each iceboat, Wtot = F s. At the finish line, each iceboat has a kinetic energy equal to the work Wtot done on it, because each iceboat started from rest. So both iceboats have the same kinetic energy at the finish line!



Example 6.6 (SOLN)

You might think this is a ‘trick’ question, but it isn’t. if you really understand the physical meanings of quantities such as kinetic energy, you can solve problems more easily and with better insight into the physics.

Notice that we didn’t need to say anything about how much time each iceboat took to reach the finish line. This is because the work-energy theorem makes no direct reference to time, only to displacement. In fact the iceboat of mass m takes less time to reach the finish line than does the larger iceboat of mass 2m; we leave the calculation to you (Exercise 6.18).




Work and Kinetic Energy in Composite Systems

• Consider a man standing on frictionless roller skates on a level surface, facing a rigid wall shown below.




Work and Kinetic Energy in Composite Systems

• There is no vertical displacement, so , , and do no work.

• also do no work as the body does not moves.

• The work done by these external forces is zero, but his kinetic energy can change nonetheless, due to various part of his body interacting.

wur

2nuur

1nur

Fur



6.3 Work and Energy with Varying Forces

• When you stretched a spring, the force you exerted is not constant as the spring is stretched.

• To illustrate, suppose a particle moves along x-direction from point to .

1x 2x




• To find the work done we divide the total displacement into small segments , , and so on.

• The work done by the force in the total displacement from to is approximately

axΔ bxΔ

1x 2x

....a a bW F x F x= Δ + Δ +

• Therefore the integral of from to : xF 1x 2x2

1

(6.7)x

xx

W F dx=∫(varying x-component of force, straight line

displacement)




• On a graph of force as a function of position, the total work done by the force is represented by the area under the curve between initial and final positions.

• When is constant, Eq. (6.7) becomes

xF

( )2 2

1 1

2 1

x x

x x xx x

W F dx F dx F x x= = = −∫ ∫




• The work done by a constant force F in the x-direction as a particle moves from to is equal to the rectangular area under the graph of force versus displacement.1x 2x




• Thus when we apply the previous concept to a spring being stretched, we have to apply a force with magnitude at each end. The following expression is obtained:

(6.8)xF kx=

F

(force required to stretch a spring)

where = force constant (or spring constant) of

spring (N/m or lb/ft)

k







• The observation that elongation is directly proportional to force for elongation that are not too great is known as Hooke’s law.

• The work done by a force when the elongation goes from zero to a maximum value X is

2

0 0

1 (6.9)

2

X X

xW F dx kx dx kX= = =∫ ∫




• The area of the shaded triangle also representing the total work done by the force.

( )( ) 21 12 2

W X kX kX= =




• If initially the spring is stretched a distance , the work done to stretched to is

1x

2x

2 2

1 1

2 22 1

1 1 (6.10)

2 2

x x

xx x

W F dx kx dx kx kx= = = −∫ ∫

• When you applied a force on the spring, the work done by you is positive.

• In contrast, the work that the spring does on you is given by negative.



Example 6.7 Work done on spring scale

A woman weighing 600N steps on a bathroom scale containing a stiff spring. In equilibrium the spring is compressed 1.0 cm under her weight. Find the force constant of the spring and the total work done on it during the compression.



Example 6.7 (SOLN)

In equilibrium the upward force exerted by the spring balances the downward force of the woman’s weight. We’ll use this principle and Eq. (6.8) to determine the force constant k, and we’ll use Eq. (6.10) to calculate the work W that the woman does on the spring to compress it. We take positive values of x to correspond to elongation, so that the displacement of the spring and the force that the woman exerts on it are both negative.

Identify and Set up



Example 6.7 (SOLN)

With our choice of coordinates, the top of the spring is displaced by x = -1.0cm = -0.010m and the force the woman exerts on the spring is Fx = -600N. Form Eq. (6.8) the force constant is

Execute

46006.0 10 /

0.010xF N

k N mx m

−= = = ×

−Then, using x1 = 0 and x2 = -0,010 m in Eq.(6.10),

2 22 1

1 1

2 2W kx kx= −

4 21(6.0 10 / )( 0.010 ) 0 3.0

2N m m J= × − − =



Example 6.7 (SOLN)

The applied force and the displacement of the end of the spring were in the same direction, so the work done must have been positive - which is just what we found by calculation. Our arbitrary choice of the positive direction has no effect on the answer for W. (You can test this by taking the positive x-direction to correspond to compression. You’ll got the same values for k and for W.)

Evaluate




Work-Energy Theorem for Straight-Line Motion, Varying Forces• An alternative derivation of the work-energy theorem is to change the

upper and lower limits of integral from to .• We note that the acceleration, a, of a particle can be expressed in

various ways through chain rulex v

(6.11)x x xx x

dv dv dvdxa v

dt dx dt dx= =




Work-Energy Theorem for Straight-Line Motion, Varying Forces• Using this results, Eq. (6.7) tells us that the total work done

by the net force is

2 2 2

1 1 1

(6.12)x x x

xtot x x x

x x x

dvW F dx ma dx mv dx

dx= = =∫ ∫ ∫

• After cancellation of dx

2

1

2 22 1

1 1 (6.13)

2 2

x

tot x xx

W mv dv mv mv= = −∫




Work-Energy Theorem for Straight-Line Motion, Varying Forces• We get the same results as Eq. (6.6) but without the assumption that

the net force F is constant.• Work-energy theorem is still valid even when F varies during

displacement.



Example 6.8 Motion with a varying force

An air-track glider of mass 0.100kg is attached to the end of a horizontal air track by a spring with force constant 20.0 N/m. initial the spring is un-stretched and the glider is moving at 1.50 m/s to the right. Find the maximum distance d that the glider moves to the right a) if the air track is turned on so that there is no friction, and b) if the air is turned off so that there is kinetic friction with coefficient .0.47kμ =



Example 6.8 (SOLN)

The force exerted by the spring is not constant, so we cannot use the constant-acceleration formulas of Chapter 2 to solve this problem. Instead, we’ll use the work-energy theorem, which involves the distance moved (our target variable) through the formula for work.

Identify



Example 6.8 (SOLN)

Figure b and c shows the FBDs for the glider without friction and with friction, respectively. We choose the positive x-direction to be the right (in the direction of the glider’s motion), with x = 0 at the glider’s initial position (where the spring is relaxed) and x = d (the target variable) at the position where the glider stops.

Set up



Example 6.8 (SOLN)

In both cases the motion is purely horizontal, so only the horizontal force work. Note that Eq. (6.10) gives the work done on the spring as it stretches, but to use the work-energy theorem we need the work done by the spring on the glider – which is the negative of Eq. (6.10).

Set up



a)As the glider moves from x1 = 0 to x2 = d, it does an amount of work on the spring given by Eq.(6.10):

The amount of work that the spring does on the glider is the negative of this value, or

The spring stretches until the glider comes instantaneously to rest, so the final kinetic energy K2 is zero.

Example 6.8 (SOLN)Execute

2 2 21 1 1(0)

2 2 2W kd k kd= − =

21

2kd−



The initial kinetic energy is , where v1 = 1.50m/s is the glider’s initial speed. Using the work-energy theorem, we find

Example 6.8 (SOLN)Execute

21

1

2mv

2 21

1 10

2 2kd mv− = −

The stretched spring subsequently pulls the glider back to the left so the glider is at rest only instantaneously.

10.100

(1.50 / )20.0 /

m kgd v m s

k N m= = 0.106 10.6m cm= =

So the distance the glider moves is



Example 6.8 (SOLN)

b) If the air is turned off, we must also include the work done by the constant force of kinetic friction. The normal force n is equal in magnitude to the weight of the glider, since the track is horizontal and there are no other vertical forces. The magnitude of the kinetic friction force is then . The friction force is directed opposite to the displacement, so the work done by friction is

Execute

k k kf n mgμ μ= =

cos180fric k k kW f d f d mgdμ= °=− =−



Example 6.8 (SOLN)

The total work is the sum of Wfric and the work done by the spring, . Hence

Execute

21

2kd

2 21

1 10

2 2k mgd kd mvμ− − = −2 21

(0.47)(0.100 )(9.81 / ) (20.0 / )2

kg m s d N m d− −

21(0.100 )(1.50 / )

2kg m s=−

2(10.0 / ) (0.461 ) (0.113 ) 0N m d N d N m+ − =g



Example 6.8 (SOLN)

This is a quadratic equation for d. The solutions are

= 0.086 m or -0.132m

We have used d as the symbol for a positive displacement, so only the positive value of d makes sense. Thus with friction the glider moves a distance

Execute

2(0.461 ) (0.461 ) 4(10.0 / )( 0.113 )

2(10.0 / )

N N N m N md

N m

− ± − −=

g

0.086 8.6d m cm= =



Example 6.8 (SOLN)

With friction present, the glider goes a shorter distance and the spring stretches less, as you might expect. Again the glider stops instantaneously, and again the spring force pulls the glider to the left; whether it moves or not depends on how great the static friction force is. How large would the coefficient of static friction have to be to keep the glider form springing back to the left?

Evaluate

sμ



Example 6.9 Motion on a curved path I

At a family picnic you are appointed to push your cousin Throcky in a swing as shown in the figure. His weight w, the length of the chains is R, and you push him until the chains make an angle with the vertical. To do this, you exert a varying horizontal force that starts at zero and gradually increases just enough so that he and the swing move very slowly and remain very nearly in equilibrium. What is the total work done on Throcky by all forces? What is the work done by the tension T in the chains? What is the work you do by exerting the force ? (Neglect the weight of the chains and seat.)

0φFur

Fur



Example 6.9 Motion on a curved path I



Example 6.9 (SOLN)

The motion is along a curve, so we will use Eq. (6.14) to calculate the work done by the tension force, by the force , and by the net force. The figure below shows the free-body diagram as well as the coordinate system. We have replaced the tension in the two chains with a single tension T.

Identify and Set up

Fur



Example 6.9 (SOLN)

Identify and Set up



Example 6.9 (SOLN)

There are two ways to find the total work done during the motion: (1) by calculating the work done by each force and then adding the quantities of work together, and (2) by calculating the work done by the net force. The second approach is easier as in this situation Throcky is in equilibrium at every point. Hence the net force on him is zero and the total work done on him by all forces is zero.

Execute



Example 6.9 (SOLN)

It’s also easy to calculate the work done by the chain tension on Throcky, because the chain tension is perpendicular to the direction of the motion at all points along the path. Hence at all points the angle between the chain tension and the displacement vector is and the scalar product in Eq. (6.14) is zero. Thus the chain tension does zero work.

Execute

dlr

90°



Example 6.9 (SOLN)

To compute the work done by , we need to know how this force varies with the angle . Throcky is in equilibrium at every point, so from we get

Execute

Fur

θ0xF =∑

( sin ) 0F T θ+ − =

and from we find0yF =∑( )cos 0T wθ + − =

By eliminating T from these two equations, we obtain

tanF w θ=



Example 6.9 (SOLN)

The point where is applied swings through the arc s. The arc length s equals the radius R of the circular path multiplied by the length (in radians), so . Therefore the displacement corresponding to a small change of angle has a magnitude . The work done by is

Execute

Fur

Fur

θ s Rθ=dθ

dl ds Rdθ= =dθ

cosW F dl F dsθ= ⋅ =∫ ∫ur r



Example 6.9 (SOLN)

Now we express everything in terms of the varying angle:

Executeθ

( ) ( )0 0

0

tan cos sin o

W w R d wR dθ θ

θ θ θ θ θ= =∫ ∫

( )01 coswR θ= −



Example 6.9 (SOLN)

If , there is no displacement; in that case, and , as we should expect. If , then and . In that case the work you do is the same as if you had lift Throcky straight up a distance R with a force equal to his weight w. In fact, the quantity is the increase in his height above the ground during the displacement, so for any value of the work done by force is the change in height multiplied by the weight. This is an example of a more general result that we’ll prove in Section 7.1.

Evaluate

0 0θ = 0cos 1θ =0W = 0 90θ = ° 0cos 0θ =

W wR=

( )01 cosR θ−

Fur



Example 6.10 Motion on a curved path II

In example 6.9 the infinitesimal vector displacement shown in the figure has a magnitude of ds, an x-component of , and a y-component of . Hence we can write . Use this expression and Eq. (6.14) to calculate the work done during the motion by the chain tension, by the force of gravity, and by the force .F

ur

cosds θ sinds θ$cos sindl ids jdsθ θ= +

r $



Example 6.10 (SOLN)

Using the same free-body from Example 6.9. The only difference is that we calculate the scalar product in Eq. (6.14) by using Eq. (1.21) for the scalar product in terms of components.

Identify and Set up



Example 6.10 (SOLN)

From the free-body diagram, we can write the three forces in terms of unit vector:

Execute

( ) $sin cosT i T jTθ θ= − +ur $

$( )w j w= −ur

$F jF=ur



Example 6.10 (SOLN)

To use Eq. (6.14), we must calculate the scalar product of each of these forces with . Using Eq. (1.21),

Execute

dlr

( )( ) ( )( )sin cos cos sin 0T dl T ds T dsθ θ θ θ⋅ = − + =ur r

( )( )sin sinw dl w ds w dsθ θ⋅ = − = −ur r

( )cos cosF dl F ds F dsθ θ⋅ = =ur r



Example 6.10 (SOLN)

Since , the integral of this quantity is zero and the work done by the chain tension is zero (just as we found in Example 6.9). Using as in Example 6.9, the work done by the force of gravity is

Execute

0T dl⋅ =ur r

ds Rdθ=

( )0

0

sin sinw dl w Rd wR dθ

θ θ θ θ⋅ = − = −∫ ∫ ∫ur r

( )01 coswR θ= − −



Example 6.10 (SOLN)

The work done by gravity is negative because gravity pull down while Throcky moves upward. Finally, the work done by the force is the integral , which we calculated in Example 6.9; the answer is .

Execute

Fur

cosF dl F dsθ⋅ =∫ ∫ur r

( )01 coswR θ+ −



Example 6.10 (SOLN)

As a check on our answers, note that the sun of all three quantities of work is zero. This is just what we concluded in Example 6.9 using the work-energy theorem.

The method of components is often the most convenient way to calculate scalar products. Use it when it will make your life easier!

Evaluate




• Suppose a particle moves from to along a curve as shown,

Fur

Work-Energy Theorem for Motion Along a Curve

where we let

= one of the many infinitesimal vector displacement (which is tangent to the path of direction). = force a a typical point along the point. = the angle between and at this point.

Fur

dlr

dlr

φ



• The small element of work done on the particle during the displacement may be written as


• When we break down the force into components, Work-Energy Theorem for Motion Along a Curve

dlrdW

cosdw F dl F dl F dlφ= = = ⋅P

ur r

where is the component of in the direction parallel to .

cosF F φ=P Fur

dlr




• The total work done by on the particle as it moves from to is then


Fur

1P 2P

2 2 2

1 1 1

cos (6.14)P P P

P P P

W F dl F dl F dlφ= = = ⋅∫ ∫ ∫P

ur r

(work done on a curved path)

• The force contributing to the work is the component of force parallel to the displacement, . cosF F φ=P




• Note that does work on the particle, the component perpendicular to the path, , has no effect on the particle’s speed; it’s only acts to change the particle’s speed.

• The integral in Eq. (6.14) is called a line integral. We usually express the line integral in terms of some scalar variable.


FPsinF F φ⊥ =



6.4 Power• In normal term, the word “power” is often relate to “energy” or “force”.• In physics, power is the time rate at which work is done.• Like work and energy, power is a scalar quantity.• Average power, , is defined when a quantity of work is done during a time interval .

avP

WΔ tΔ

(6.15)avW

Pt

Δ=Δ

(average power)



6.4 Power

• When the rate of work done is not constant, we can define it as instantaneous power, P.

0lim (6.16)t

W dWP

t dtΔ →

Δ= =Δ

(instantaneous power)

• SI unit is watt (W).

3

6

1 W = 1 J/s

1 kW = 10 W

1 MW = 10 W



• For British unit, it is expressed as foot-pound per second ( ).

• A larger unit is called horsepower (hp).

6.4 Power

• For usual commercial unit of electrical energy, we use kilowatt-hour (kWh).

• The kilowatt-hour is a unit of work or energy, not power.

ft lb/s⋅

1 hp = 550 ft lb/s = 33,000 ft lb/min⋅ ⋅

1 hp = 746 W = 0.746 kW

( )( )3 61 kWh = 10 J/s 3600 s = 3.6 10 J = 3.6 MJ×



• In mechanics we can express power in terms of force and velocity.

• Suppose that a force acts on a body while it undergoes a vector displacement .

• If is the component of tangent to the path (parallel to ), then the work done by the force is , and the average power is

6.4 Power

Fur

sr

FP Fur

sΔr

W F sΔ = ΔP

(6.18)av avF s s

P F F vt t

Δ Δ= = =Δ ΔP

P P



• Instantaneous power P is the limit of this expression :

6.4 Power

0tΔ →

(6.18)P F v= P

where = magnitude of the instantaneous velocity. v

• We can also express Eq. (6.18) in terms of scalar product:

(6.19)P F v= ⋅ur r

(instantaneous rate at which force does work on a particle)

Fur



Example 6.11 Force and Power

Each of the 2 jet engines on a Boeing 767 airliner develops a thrust (a forward force on the airplane) of 197,000 N (44,300 lb). When the airplane is flying at 250 m/s (900 km/h, or roughly 560 mi/h), what horsepower each engine develop?



Example 6.11 (SOLN)

The thrust is in the direction of motion, so in Eq. (6.18) is just equal to the thrust. Our target variable is the instantaneous power P.

Identify and Set up

sF

Execute

At , each engine develops power P given by Eq. (6.18):

250 m/sv =

( )( )5 71.97 10 N 250 m/s 4.93 10 WsP F v= = × = ×

( )7 1 hp4.93 10 W 66,000 hp

746 W= × =



Example 6.11 (SOLN)

The speed of modern airliners is directly related to the power of their engines. The largest engines on propeller-driven airliners of the 1950s developed about 3400 hp , and these airliners had maximum speeds of about 660 km/h (370 mi/h). The power of each of the engines of a Boeing 767 is nearly 20 times greater, enabling it to fly at about 900 km/h (560 mi/h) and to carry a much heavier load.

Evaluate

( )62.5 10 W×



Example 6.11 (SOLN)

What if the engine are running at maximum thrust while the airliner is at rest on the ground (with brakes held)? Because , the engines develop zero power. Force and power are not the same thing!

Evaluate

0v =



Example 6.12 Force and Power

As part of a charity fund-raising drive, a Chicago marathon runner with mass 50.0 kg runs up the stair to the stairs to the top of the 443 m tall tower. To lift herself to the top in 15.0 min, what must be her average power output in watts? In kilowatts? In horsepower?



Example 6.12 (SOLN)

We’ll treat runner as a particle of mass m. we can calculate her average power output in two ways: (1) by first determining how much work she must do and then dividing by the elapsed time, as in Eq. (6.15) or (2) by calculating the average upward force she must exert (in the direction of the climb) and multiplying it by her upward velocity, as in Eq (6.17).

Identify and Set up



Example 6.12 (SOLN)

As in Example 6.9, lifting a mass m against gravity requires an amount of work equal to the weight mg multiplied by the height h it is lifted. Hence the work she must do is

Execute

( )( )( )250.0 kg 9.80 m/s 443 mW mgh= =

The time is 15.0 min = 900 s, so from Eq. (6.15) the average power is

52.17 10 J241 kW=0.323 hp

900 savP×= =



Example 6.12 (SOLN)

Let’s try the calculation again using the alternative approach with Eq. (6.17). The force exerted is vertical, and the average vertical component of velocity is , so the average power is

Execute

( ) ( )433 m / 900 s 0.492 m/s=

( )av av avP F v mg v= =P

( )( )250.0 kg 9.80 m/s=

which is the same result as before.



Example 6.12 (SOLN)

In fact the runner’s total power output will be several times greater than we have calculated. The reason is that the runner isn’t really a particle but a collection of parts that exert forces on each other and do work, such as the work done to inhale and exhale and to make her arms and legs swing. What we’ve calculated is only the part of her power output that goes into lifting her to the top of the building.

Evaluate



• When a constant force acts on a particle that undergoes a straight-line displacement , the work done by the force on the particle is defined to be the scalar product of and .

The unit of work in SI units is 1 joule = 1 newton-meter (1 J = 1 N·m) . Work is a scalar quantity; it has an algebraic sign (positive or negative) but no direction in space.

Concept Summary

Fur

Fur

sr

sr



• The kinetic energy K of a particle equals the amount of work required to accelerate the particle from rest to speed v. It is also equal to the amount of work the particle can do in the process of being brought to rest. Kinetic energy is a scalar quantity that has no direction in space; it is always positive or zero. Its units are the same as the units of work:

Concept Summary

2 21 J = 1 N m = 1 kg m /s⋅ ⋅



• When forces act on a particle while it undergoes a displacement, the particle’s kinetic energy changes by an amount equal to the total work done on the particle by all the forces. This relations, called the work-energy theorem, is valid whether the forces are constant or varying and whether the particle moves along a straight line or curved path. It is applicable only to the bodies that can be treated as a particle.

Concept Summary



• When a force varies during a straight-line displacement, the work done by the force is given by an integral in Eq. (6.7).• If the particle follows a curved path, then the work done by a force is given by an integral that involves the angle between the force and the displacement.

This expression is valid even if the force magnitude and the angle vary during the displacement.

Concept Summary

θFur



• Power is the time rate of doing work. The average power is the amount of work done in time divided by that time. The instantaneous power is the limit of the average power as goes to zero. When a force acts on a particle moving with a velocity , the instantaneous power (rate at which the force does work) is the scalar product of and . Like work and kinetic energy, power is a scalar quantity. The SI unit of power is 1 watt = 1 joule/second (1 W = 1 J/s).

Concept Summary

Fur



Key Equations

cosW F s Fs φ= ⋅ =ur r

angle between and (6.2), (6.3)F sφ =ur r

21 (6.5)

2K mv=

2 1 (6.6)totW K K= −

2

1

(6.7)x

xx

W F dx=∫



Key Equations

2 2

1 1

cos P P

P P

W F dl F dlφ= =∫ ∫ P

2

1

(6.14)P

P

F dl= ⋅∫ur r



Key Equations

(6.15)avW

Pt

Δ=Δ

0lim (6.16)

W dWP

t dtΔ→

Δ= =Δ

(6.19)P F v= ⋅ur r

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Module 7Work and KE