Physics 2210

Homework 8

Spring 2015

Charles Jui

February 23, 2015

IE Block Spring Incline

Wording

A 5 kg block is placed near the top of a frictionless ramp, which makes an angleof 30◦ degrees to the horizontal. A distance d = 1.3 m away from the block isan unstretched spring with k = 3× 103 N/m. The block slides down the rampand compresses the spring. Find the maximum compression of the spring.See Figure 1.

Figure 1: IE Block Spring Incline

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Solution

Since only conservative forces are at play, the total mechanical energy is con-served 1, i.e. MEi = MEf

mg(xmax + d) sin θ =1

2kx2

max

which yields quadratic equation in xmax

x2max − 2mg

kxmax − 2mgd

k= 0

and its solutions are

x±max =

mg sin θ

k

(1±

√1 +

2kd

mg sin θ

)

We eliminate the negative solution since it would mean that the block is not incontact with the spring. And we will get

xmax = 0.1541949m

Tipler6 7.P.012.

Wording

Estimate the following:

(a) The change in your gravitational potential energy on taking an elevator fromthe ground floor to the top of the Empire State Building. The buildingis 102 stories high (assuming a 3 min ride to the top of the building).(Assuming your mass is 68 kg and the height of one story to be 3.5 m.)

(b) The average force exerted by the elevator on you during the trip.

(c) The average power delivered by that force.

Solution

(a)

Change in potential energy is given by

∆U = mg∆h

where ∆h = 102× 3.5m = 357m, thus

∆U = 238.14756 kJ

1Note adding xmax in the potential energy.

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(b)

The average force is defined by

F̄ =1

h2 − h1

∫ h2

h1

F · dx

but ∆h = h2 − h1 and the integral is simply the work done, which in this caseis equal to the change in potential energy, i.e.

F̄ =∆U

∆h

So we get

F̄ = 667.08N

(c)

The average power is defined as

P̄ =Work

time=

∆U

∆t

where the time is t = 3.5× 60 s = 210 s, so

P̄ = 1.323042 kW

Tipler6 7.P.034.

Wording

A simple Atwood’s machine uses two masses, m1 and m2. Starting from rest,the speed of the two masses is 3 m/s at the end of 7 s. At that instant, thekinetic energy of the system is 54 J and each mass has moved a distance of 10.5m. Determine the values of m1 and m2. See Figure 2.

Solution

Here we have conservation of energy: At the beginning, the total energy iszero.2 After a while

0 =1

2m1v

21 +

1

2m2v

22 +m1gh1 +m2gh2

Let’s assume that the first mass goes down; now we have

h2 = −h1 = d v1 = −v2 = v

Hence we obtain

0 =1

2(m1 +m2)v

2 + (m2 −m1)gd

2We can pick the masses to be at the same height, for convenience.

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Figure 2: Tipler6 7.P.034.

We are also given the total kinetic energy, i,e,

KE =1

2m1v

21 +

1

2m2v

22 =

1

2(m1 +m2)v

2

Combining the two equations produces the following system

2KE = (m1 +m2)v2

KE = (m1 −m2)gd

Solving this system yields

(a)

m1 =

(1

v2+

1

2gd

)KE

m1 = 6.262 kg

(b)

m2 =

(1

v2− 1

2gd

)KE

m2 = 5.738 kg

Tipler6 7.P.036.

Wording

A block of mass m is pushed up against a spring, compressing it a distance x,and the block is then released. The spring projects the block along a friction-less horizontal surface, giving the block a speed v. The same spring projects a

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second block of mass 3m, giving it a speed of 4v. What distance was the springcompressed in the second case? Give your answer in terms of a numerical factortimes x, the compression of the first block.

Solution

Transfer of energy in the first case

1

2kx2 =

1

2m1v

21

Transfer of energy in the second case

1

2ky2 =

1

2m2v

22

Taking ratio of the two equations yields

y2

x2=

m2v22

m1v21=

m2

m1· v

22

v21

Thus

y =

√3m

m

4v

vx = 4

√3x

y = 6.928x

Tipler6 7.P.060.

Wording

In a volcanic eruption, 2km3 of mountain with a density of 1600 kg/m3 waslifted an average height of 513m.

(a) How much energy in joules was released in this eruption?

(b) The energy released by thermonuclear bombs is measured in megatons ofTNT, where 1 megaton of TNT = 4.2× 1015 J . Convert your answer for(a) to megatons of TNT.

Solution

(a)

The energy required is equal to the change of potential energy of the ash, i.e.

E = ∆U = Mg∆h = ρV g∆h

where V = 2× 109 m3. This gives value

E = 1.610E + 16 J

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(b)

Simple ratio E/(4.2× 1015 J) yields

3.834 Mton TNT

Pendulum

Wording

A mass m = 4.7 kg hangs on the end of a massless rope L = 2.05 m long. Thependulum is held horizontal and released from rest.

1) How fast is the mass moving at the bottom of its path?

2) What is the magnitude of the tension in the string at the bottom of thepath?

3) If the maximum tension the string can take without breaking is Tmax =382N , what is the maximum mass that can be used? (Assuming that themass is still released from the horizontal and swings down to its lowestpoint.)

4) Now a peg is placed 4/5 of the way down the pendulum’s path so that whenthe mass falls to its vertical position it hits and wraps around the peg. Asit wraps around the peg and attains its maximum height it ends a distanceof 3/5 L below its starting point (or 2/5 L from its lowest point). How fastis the mass moving at the top of its new path (directly above the peg)?

5) Using the original mass of m = 4.7 kg, what is the magnitude of the tensionin the string at the top of the new path (directly above the peg)?

Figure 3: Pendulum 1

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Figure 4: Pendulum 2

Solution

1)

Forces acting on the pendulum are conservative, thus the energy is conserved,i.e.

mgL =1

2mv2

so we getv =

√2gL

and numerically

v = 6.342m/s

2)

At the bottom, the vertical component of the acceleration must be equal tocentripetal acceleration. At the same time, the net force is made up of gravityand tension; hence according to the II.Newton law

T −mg = mv2

L

which gives3

T = m

(g +

v2

L

)= m

(g +

2gL

L

)= 3mg

and thus

T = 138.321N

3)

Turning around the result of the previous part

mmax =Tmax

3g

mmax = 12.980 kg

3Using results of the previous part for the speed at the bottom.

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4)

Once again, all the force involved are conservative, thus the energy is conserved

mghi = mghf +1

2mv′2

so

v′ =√2g(hi − hf ) =

√2g(3/5)L =

√(6/5)gL

which gives

v′ = 4.912m/s

5)

Here the approach is analogous to the one in part 3) with the modification ofthe net force:

T ′ +mg = mv′2

(1/5)L

T ′ = m

(5v‘2

L− g

)= 5mg

with the value of

T ′ = 230.535N

Loop the Loop

Wording

A mass m = 73 kg slides on a friction-less track that has a drop, followed by aloop-the-loop with radius R = 18.1 m and finally a flat straight section at thesame height as the center of the loop (18.1 m off the ground). Since the masswould not make it around the loop if released from the height of the top of theloop (do you know why?) it must be released above the top of the loop-the-loopheight. See the Figure 5. (Assume the mass never leaves the smooth track atany point on its path.)

Figure 5: Loop the Loop

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1) What is the minimum speed the block must have at the top of the loop tomake it around the loop-the-loop without leaving the track?

2) What height above the ground must the mass begin to make it around theloop-the-loop?

3) If the mass has just enough speed to make it around the loop without leavingthe track, what will its speed be at the bottom of the loop?

4) If the mass has just enough speed to make it around the loop without leavingthe track, what is its speed at the final flat level (18.1 m off the ground)?

5) Now a spring with spring constant k = 15300 N/m is used on the final flatsurface to stop the mass. How far does the spring compress?

6) It turns out the engineers designing the loop-the-loop didn’t really knowphysics – when they made the ride, the first drop was only as high as thetop of the loop-the-loop. To account for the mistake, they decided to givethe mass an initial velocity right at the beginning. How fast do they needto push the mass at the beginning (now at a height equal to the top of theloop-the-loop) to get the mass around the loop-the-loop without fallingoff the track?

Solution

1)

At the top of the loop, the centripetal force needs to be at least as large as theforce of gravity, hence

Fg = mv2min

Rso we get

vmin =√

gR

and numerically

vmin = 13.325m/s

2)

Requirement of the previous part puts the total energy of the box at the top ofthe loop to

TE =1

2mv2min +mg(2R)

which needs to be equal to the original potential energy

mgh =1

2mv2min +mg(2R)

thus getting

h = 2R+v2min

2g= 2R+ (1/2)R =

5

2R

hence

h = 45.25m

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3)

So energy conservation dictates

mg5

2R =

1

2mv2b

Solving for the speedvb =

√5gR

and

vb = 29.796m/s

4)

mg5

2R =

1

2mv2f +mgR

givesvf =

√3gR

so

vf = 23.080m/s

5)

The kinetic energy will be transferred to the spring

1

2mv2f =

1

2kx2

Solving for distance

x =

√mv2fk

=

√3mgR

k

hence

x = 1.594m

6)

So the initial energy is

TEi =1

2mv20 +mg(2R)

which must be equal to the energy at the top of the loop

TE =1

2mv2min +mg(2R)

Solving for the initial velocityv0 = vmin

or

v0 = 13.325m/s

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