PHY 303K Test 2 Solutions

Version 058 Test 2 florin (57850) 1

This print-out should have 20 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.

001 10.0 points

A tennis ball at the end of a string is re-leased from rest and swings through a verticalposition, reaching its maximum speed there,then comes to momentary rest on the otherside.

rest rest

What is the direction of the balls accel-eration when it is at its lowest point and atmaximum speed?

1. correct

2. 3.

4.

5.

6.

Explanation:

At the lowest pointd v

dt= 0 so the entire

acceleration is ar and is directed toward thecenter of the circle.

002 10.0 points

Suppose the potential energy of an objectis given by:

U = a r + b r3

where a = 10 J/m2 and b = 7.4 J/m4. What isthe force on the object when it is at r = 3.8 m?1. -530.02. -82.0843. -223.7044. -34.7755. -330.5686. -195.167. -141.3178. -24.1169. -127.21710. -631.0

Correct answer: 330.568 N.Explanation:

The force exerted on an object is given byF = dU/dr so taking the derivative of thepotential energy function gives

F = dU/dr = (a+ 3b r2)and inserting r = 3.8 m gives

F = [(10 J/m2) + 3(7.4 J/m4)(3.8 m)2]

F = 330.568 N

003 10.0 points

A satellite in a low circular earth orbit issubjected to a very small constant frictionalforce f , due to the resistance of the thin atmo-sphere at that altitude. As the satellite losesenergy, the radius of the orbit r slowly de-creases, and the satellite spirals downwards.Calculate the change in total energy E of

the satellite per revolution. Assume the orbitis approximately circular over a revolution.

1. E = 4 f r3/3 pi

2. E = 2 pi r f

3. E = f2/r2

4. E = f/2 pi r

5. E = 2 pi r f correct

6. E = 3 pi2 r f


7. E = f

8. E = f/2 pi r

9. E = f r

10. E = pi2 fExplanation:

The energy lost per revolution is just thework done by the frictional force along theapproximately circular orbit of radius r. Weknow that the frictional force acts in a direc-tion opposite to that of instantaneous direc-tion of motion, so the work done by this non-conservative force is Wnc = f s = 2 pi r f ,so from conservation of energy,

E = Ef Ei =Wnc = f (2 pi r) .

004 10.0 points

Imagine a thin tower 4000 miles tall; .e.as tall as the earths own radius Re, placedat the north pole of the earth. Suppose youstart at rest at the base of the tower and climbto the top (wearing a spacesuit and carryingsupplies you need, so that your total mass isM).How much work would you have done by

the time you are at the top of the tower andat rest again? Neglect the mass of the towercompared to the mass of the earth.

1. W = GMeM Re

2. W =GMeM

2Recorrect

3. W =2GMeM

5Re

4. W =GMeM

3Re

5. W = 4GMeM Re

6. W =GMeM

4Re

7. W =3GMeM

2Re

8. W = 2GMeM Re

9. W =5GMeM

2ReExplanation:

If we consider the system to be you plus theearth,the gravitational potential energy is

Ug(r) = GMeMr

.

The initial energy is Ei = Ug,i and thefinal energy is Ef = Ug,f . Thus the EnergyPrinciple gives:

W = Ug = Ug (2Re) Ug (Re) = GMeM2Re

.

Alternately, if you consider the system tojust be you, the net work is zero and both youand the earth do work on the system.

W =Wg =

2RERE

GMEm

r2dr

005 10.0 points

A 12 g bullet is accelerated in a rifle barrel76.5 cm long to a speed of 455 m/s.Use the work-energy theorem to find the

average force exerted on the bullet while it isbeing accelerated.1. 7025.872. 1623.733. 3047.764. 6173.345. 1029.26. 1676.067. 1749.518. 7786.289. 4034.8710. 1512.17

Correct answer: 1623.73 N.

Explanation:

Let : m = 12 g = 0.012 kg ,

x = 76.5 cm , and

v = 455 m/s .


The kinetic energy of the bullet before itis fired is 0 J since it is not moving. Atthe moment that it leaves the rifle barrel, itskinetic energy is

K =1

2mv2

=1

2(0.012 kg) (455 m/s)2

= 1242.15 J .

Since the work done on the bullet is equal toits change in kinetic energy, this is also thework that is done on the bullet:

W =1

2mv2 = 1242.15 J .

Also, the work is given by

W = Favg x

Favg =W

x=

1242.15 J

0.765 m

= 1623.73 N .

006 10.0 points

There was once a plan to roll cars partwaythrough the English Channel and tow themthe rest of the way. Imagine that a car of massm = 3938 kg rolls down a slope of length` = 51 km and at an angle = 6.8 tothe horizontal as shown in the picture. Alsoimaging there is a small amount of frictionon the car as it rolls given by coefficient offriction k = 0.1.

What would the speed of the car be whenit reaches the level part of the track, if the carstarts from rest? The acceleration of gravityis 9.8 m/s2 .1. 225.4072. 148.6223. 242.2474. 145.0585. 258.8246. 233.491

7. 138.2028. 96.33939. 217.33910. 163.76

Correct answer: 138.202 m/s.

Explanation:

Let : m = 3938 kg ,

g = 9.8 m/s2 ,

` = 51 km = 51000 m , and

= 6.8 .

The change in potential energy is

U = Uf Ui= 0mg y= mg ` sin = (3938 kg) (9.8 m/s2) (51 km) sin 6.8= 2.33044 108 J

with a magnitude of 2.33044 108 J .

Let : k = 0.1 .

From the Energy Principle, while rollingdownhill we have

K +U =Wsurrmv2

2= U +Wf

v =

2 (U Wf )

m.

Since

U W = U kmg ` cos = 2.33044 108 J 0.1(3938 kg)(9.8 m/s2) (51000 m) cos 6.8

= 3.76074 107 J , then

v =

2(3.76074 107 J)

3938 kg= 138.202 m/s .


007 10.0 points

A little boy and little girl in a playground aresitting on a large metal disk which is rotatingfreely at a constant number of revolutions perminute. The boy is at the outer rim, and thegirl is about halfway between the rim and thecenter.How do their accelerations compare?

1. Both boy and girl have the same acceler-ation.

2. The boys acceleration is greater. cor-rect

3. Neither is accelerating since the disk hasa constant speed of rotation.

4. The girls acceleration is greater.

5. None of these

Explanation:

The speed of a point on a rotating disk isproportional to the distance from the center ofthe disk: v r. But the acceleration towardthe center of the disc is determined by the

quantityv2

r r in this case. Thus the boys

acceleration is about twice that of the girl.

008 10.0 points

Consider a car of mass m moving withoutslipping on a circular track of radius R. Thetrack is inclined at an angle with respect tothe horizontal direction. Suppose the trackhas a coefficient of static friction s. Whatis the minimum speed vmin at which the carmust move to avoid slipping up or down thetrack?

1. vmin =

g R

cos s sin sin + s cos

2. vmin =

g R

cos s sin sin s cos

3. vmin =

g R

cos + s sin

sin + s cos

4. vmin = 0

5. vmin =

g R

sin + s cos

cos s sin

6. vmin =

g R

sin s cos cos + s sin

correct

7. vmin =

g R

cos + s sin

sin s cos 8. vmin =

9. vmin =

g R

sin s cos cos s sin

10. vmin =

g R

sin + s cos

cos + s sin

Explanation:

Call the horizontal axis the x-axis andchoose it to be the radially inward directionat a certain instant of time. Similarly call thevertical axis the y-axis and choose it to be up-ward. Therefore, the car experiences a netcentripetal acceleration along the x-axis andzero net acceleration along the y-axis. Look-ing for the minimum velocity implies thatfriction is directed up the track - if the speedwas zero the car would tend to slide down thetrack and static friction opposes the directionthat motion would tend.From The Momentum Principle,

Fy = N cos + sN sin mg = 0

N =mg

cos + s sin

Fx = N sin sN cos = mv2min

R

N (sin s cos ) = mv2min

R.Substituting the expression for N from the

y-equation gives,

mv2minR

= mgsin s cos cos + s sin


Finally, solving for vmin gives

vmin =

g R

sin s cos cos + s sin

009 10.0 points

An ice cube is held at a depth h by a stringattached to the bottom of a bucket of water.

h

As time passes the ice melts. Suppose ini-tially the ice has a volume V and the magni-tude of tension in the string is T . When thevolume of the ice has decreased to V = 1/3V ,what is the magnitude of tension T ?

1. T = 3T

2. T = 2/3T

3. T = 5/6T

4. T = 1/3T correct

5. T = T

6. T = 3/2T

7. T = 3/4T

8. T = 1/6T

9. T = 2T

10. T = 5/3T

Explanation:

First create a free-body diagram of the sce-nario. From this it is evident there are threeforces which act on the ice: the buoyant forceF

B(which acts upward), the weightW of the

ice (which of course acts downward), and thetension T of the rope (which pulls downward

on the ice). Then mathematicallyF = ma

FB T W = ma .

The ice has no acceleration and so theright-side of the above equation becomes zero.Plugging-in the formulas for both the buoyantforce and the weight, gives

T = w V g i V g= V (w i) g

The same analysis works for after the icemelts. Notice that T is proportional to V soputting T in for T and V in for V gives

T = V (w i) g= 1/3V (w i) g= 1/3T

010 10.0 points

A block of styrofoam with mass m has a leadweight with mass 1/2m glued on top. Whenthe lead-styrofoam piece is placed in water,the styrofoam piece is partially submerged inthe water. The water has a density w, andthe density of the styrofoam s = 1/3 w. Ifthe styrofoam has a total volume V , what isthe fraction of the submerged volume Vsub tothe total volume V ?

1. Vsub/V = 1

2. Vsub/V = 1/2 correct

3. cannot be determined without more in-formation

4. Vsub/V = 3/2

5. Vsub/V = 1/3

6. Vsub/V = 3/4

7. depends on the density of lead

8. Vsub/V = 2/3


9. Vsub/V = 0 by the Momentum Principle

10. Vsub/V = 1/4

Explanation:

If we consider the system to be the leadand styrofoam combination, the total mass is3/2m, which can be expressed in terms of thedensity of styrofoam as 3/2 sV , where V isthe volume of the styrofoam block.The buoyant force depends only on the vol-

ume displaced by the object and the densityof the fluid, i.e.,

FB = wVsub g

The forces on our system are just the grav-itational force and the buoyant force, whichmust sum to zero since the system is not mov-ing

3/2mg FB = 03/2mg = FB

3/2 sV g = w Vsub g

Vsub/V = (3/2) (s/w)

Vsub/V = 1/2

011 10.0 points

The figure shows four situations - one inwhich an initially stationary block is droppedand three in which the block is allowed toslide down frictionless ramps again beginningfrom rest. Rank the situations according tothe speed of the block at y = 0 from fastest toslowest.

1. v1 = v2 = v3 > v4

2. v1 > v2 = v3 > v4

3. v1 = v2 = v3 = v4 correct

4. Cannot be determined with knowing therelative angles of the ramps.

5. v1 > v2 > v3 = v4

6. v4 > v3 > v2 > v1

7. v4 > v1 = v2 = v4

8. v1 > v2 > v3 > v4

Explanation:

Using a system that includes the blockand the Earth, there is no work done bythe surroundings so the initial energy inall four cases is all potential energy withU = mg h and the final energy is all kineticthen 1/2mv2f = mg h. Thus the kinetic ener-gies and speeds of all four blocks are the samewhen they are at y = 0.

012 10.0 points

A small metal ball is suspended from the ceil-ing by a thread of negligible mass. The ballis then set in motion in a horizontal circle sothat the threads trajectory describes a cone.The acceleration of gravity is 9.8 m/s2 .

vr

g

`

m

What is the angular frequency of the ballwhen it is in circular motion? Answer in termsof g, ` and

1. =

g sin2 /`

2. =g/` tan

3. =g/` cos

4. =g/` sin correct


5. =g/` tan2

6. =g cos /`

7. =g tan /`

8. =g sin /`

9. =g/` cos2

10. =

g/` sin2

Explanation:

Use the free body diagram below.

mg

T

The tension on the string can be decom-posed into a vertical component which bal-ances the weight of the ball and a horizontalcomponent which causes the centripetal ac-celeration, acentrip that keeps the ball on itshorizontal circular path at radius r = ` sin .If T is the magnitude of the tension in the

string, then

Tvertical = T sin = mg (1)

andThoriz = macentrip

or

T cos =mv2ball` cos

. (2)

Solving (1) for T yields

T =mg

sin (3)

and substituting (3) into (2) gives

mg cos

sin =mv2ball` cos

.

Therefore,

v =

g ` cos2

sin .

and solving for yields

= v/r

=

g ` cos2 sin

` cos

=

g

` sin .

013 10.0 points

Two blocks of the same size, shape and massm are one atop the other. A constant hori-zontal force F is applied to the lower block.The coefficient of kinetic friction between thelower block and the level tabletop on whichthe blocks slide is k, and the coefficient ofstatic friction between the two blocks is s

F

k

s

m

m

1

2

Find F the magnitude of the maximumforce that can be applied so that the blocksaccelerate together.

1. F = mg (2s k)

2. F = mg (2s + k)

3. F = mg (s + 2k)

4. F = mg (s 2k)

5. F = 1/2 mg (s + k)

6. F = 2mg (k s)

7. F = 2mg (s + k) correct


8. F = 1/2 mg (k s)Explanation:

a a

fs

F

fs fk

1 2

mm

The only force on the top block is staticfriction therefore the maximum value for mais given by ma = smg Thus it must alsobe true that the maximum value for F willsatisfy F fs fk = smg. Solving for Fgives F = 2mg (s + k)

014 10.0 points

A car of mass 519 kg goes over a hill withradius 138 m at the top.What is the maximum speed that the car

can go over the hill without leaving the road?The acceleration of gravity is 9.8 m/s2 .1. 27.50562. 35.13973. 41.41264. 37.69625. 29.13216. 38.21267. 36.7758. 40.21199. 25.4910. 31.7711

Correct answer: 36.775 m/s.

Explanation:

mv2

r= mg N

where N is the normal force acting on thecar from the ground. The car will fly off theground just when N = 0 so the maximumspeed allowed will be

vmax =g r

=(9.8 m/s2)(138 m)

= 36.775 m/s .

015 10.0 points

An outfielder throws a 1.23 kg baseball at aspeed of 40 m/s and an initial angle of 11.2.What is the kinetic energy of the ball at the

highest point of its motion?1. 1743.752. 946.8773. 1853.754. 2286.235. 8689.736. 3703.397. 2026.068. 1088.829. 1068.8510. 1896.83

Correct answer: 946.877 J.

Explanation:

Given : m = 1.23 kg ,

= 11.2 , and

v = 40 m/s .

At its highest point, the baseball has only ahorizontal velocity vx = v cos contributingto its kinetic energy, so the kinetic energy is

K =1

2mv2x

=1

2m (v cos )2

=1

2(1.23 kg) [(40 m/s) cos 11.2]2

= 946.877 J .

016 10.0 points

Consider a coin which is tossed straight upinto the air. After it is released it movesupward, reaches its highest point and fallsback down again.What is its acceleration when the coin is

at its highest point, i.e. when its velocity iszero? Take up to be the positive direction.

1. The acceleration is in the negative direc-tion and constant. correct


2. The acceleration is zero.

3. The acceleration is in the positive direc-tion and increasing.

4. The acceleration is in the negative direc-tion and increasing.

5. The acceleration is in the negative direc-tion and decreasing.

6. The acceleration is in the positive direc-tion and decreasing.

7. The acceleration is in the positive direc-tion and constant.

Explanation:

Gravity supplies the acceleration so it is aconstant negative value.

017 10.0 points

A block of mass 2m has a velocity vectorgiven by ~v = v0 < 3, 4,2 >. What is K thekinetic energy of the mass?

1. K = 29mv20 correct

2. K = 22mv20

3. K = mv20 < 18, 32,8 >

4. K = 54mv20

5. K = mv20 < 18, 32, 8 >

6. K = 44mv20

7. K = mv20 < 9, 16, 4 >

8. K = 50mv20

9. K = 25mv20

10. K = mv20 < 9, 16,4 >Explanation:

IfM is the mass and~v is the velocity, kineticenergy is given by K = 1/2M (v2x + v

2

y + v2

z)

which in this case is K = (1/2) 2mv20 ((3)2 +

(4)2 + (2)2) so K = 29mv20018 10.0 points

A metal sphere of radius r = 0.12 m isdropped into the ocean and sinks to a depthof d = 3510 m .What is the total force exerted on the out-

side of the sphere at this depth? You mayassume the pressure is constant at all pointsaround the sphere. Atmospheric pressure is1.013 105 Pa, the density of seawater is = 1024 kg/m3 and the acceleration dueto gravity is g = 9.8 m/s2.1. 11299400.02. 26440600.03. 5930400.04. 6392230.05. 7182320.06. 16113900.07. 8909660.08. 23535300.09. 22524000.010. 19013000.0

Correct answer: 6.39223 106 N.Explanation:

Let : r = 0.12 m ,

d = 3510 m

= 1024 kg/m3 , and

g = 9.8 m/s2 .

The total pressure on the sphere is due to boththe atmospheric and the hydrostatic pressureof the water:

P = P0 + g d

= 1.013 105 Pa+(1024 kg/m3)(9.8 m/s2)(3510 m)

= 3.53249 107 N/m2

Now that we have the pressure we can easilycalculate the total force on the outside of thesphere as the product of the pressure timesthe surface area of the sphere A, where

A = 4pi r2


F = P A= (3.53249 107 N/m2)(4pi)(0.12 m)2= 6.39223 106 N

019 10.0 points

What is the minimum kinetic energy neededto launch a payload of mass m to an altitudethat is one Earth radius, RE, above the sur-face of the Earth (the payload will then fallback to Earth)? (Note thatME is the mass ofthe Earth.)

1. 0.5GmMERE

correct

2.GmMERE

3. 2GmMERE

4. 0.25GmMERE

5. 0.1GmMERE

Explanation:

Energy is conserved:

Ki + Ui = Kf + Uf

The kinetic energy after the payload hasbeen launched to an altitude RE above theEarths surface is

Kf = 0 .

The potential energy at Earths surface beforethe launch is

Ui = GMEmRE

,

and the potential energy at altitude RE abovethe Earths surface after the launch is

Ui = GMEm2RE

.

So, the minimum energy needed to launchthe payload to RE above Earths surface is

Ki =GMEm

RE GMEm

2RE=GmME2RE

020 10.0 points

A block of mass m is pushed a distance D upan inclined plane by a horizontal force F . Theplane is inclined at an angle with respect tothe horizontal. The block starts from rest andhas velocity is vf when the block is at D.

m

D

kF

The coefficient of kinetic friction k is

1. k =mgD sin + 1/2mv2f F D cos

mgD cos

2. k =F D cos mg sin mg cos + F sin

3. k =F cos mg sin 1/2mv2f

mg cos + F sin

4. k =F D cos mgD cos 1/2mv2f

mgD sin + F D cos

5. k =F D sin mg cos 1/2mv2f

mg sin

6. k =F D sin mg sin 1/2mv2f

mg cos

7. k =F D cos mgD sin 1/2mv2f

mgD cos + F D sin correct

8. k =F D cos mgD sin 1/2mv2f

mgD cos

Explanation:

The force of friction has a magnitudeFfriction = kN . Since it is in the direc-tion opposite to the motion, we get

Wfriction = FfrictionD= kN D.= k (mg cos + F sin )D


The normal force makes an angle of 90

with the displacement, so the work done by itis zero.The work done by gravity is

Wgrav = mgD sin .The work done by the force F is

WF = F D cos .

From the Energy Principle we know thatWnet = K ,

WF +Wgrav +Wfriction =1

2mv2f .

Thus

muk =F D cos mgD sin 1/2mv2f

mgD cos + F D sin

PHY 303K Test 2 Solutions

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