HW 13 Solutions

7/29/2019 HW 13 Solutions

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goldhagen (dfg358) HW-13 florin (57850)

This print-out should have 30 questions.Multiple-choice questions may continue onthe next column or page find all choicesbefore answering.

001 10.0 points

A 5.7 kg mass is attached to a light cord thatpasses over a massless, frictionless pulley. Theother end of the cord is attached to a 2.6 kgmass.

5.6 m

5.7 kg

2.6 kg

Use conservation of energy to determinethe final speed of the masses after the heaviermass has fallen (starting from rest) 5.6 m .Theacceleration of gravity is 9.8 m/s2 .

Correct answer: 6.40271 m/s.Explanation:

Let : m1 = 5.7 kg ,

m2 = 2.6 kg , and

= 5.6 m .

Consider the free body diagrams

5.7 kg 2.6 kg

T T

m1

g

m2

g

a a

Let the figure represent the initial config-uration of the pulley system (before m1 fallsdown).

From conservation of energy

Ki + Ui = Kf + Uf

0 + m1 g = m2 g +1

2m1 v

2 +1

2m2 v

2

(m1

m2) g =

1

2

(m1 + m2) v2

Therefore

v =

(m1 m2)(m1 + m2)

2 g

=

5.7 kg 2.6 kg5.7 kg + 2.6 kg

2 (9.8 m/s2)(5.6 m)

= 6.40271 m/s .

keywords:

002 (part 1 of 2) 10.0 pointsCalculate the speed of a satellite in a circular orbit near the Earth (just above thatmosphere). Height of the atmosphere i 50000 m. Mass and radius of the earth ar6

1024 kg and 6.37

106 m, respectively. G

= 6.7 1011 N m2kg2

.

Correct answer: 7913.08 m/s.

Explanation:Since the satellites circular orbit is entirely

caused by the gravitational force of the earthwe set the gravitational force equal to thcentripetal force to get:

mv2orbit

R=

GMm

R2

vorbit =

GM

R

=

(G)(6 1024 kg)

6.37 106 m + 50000 m= 7913.08 m/s

003 (part 2 of 2) 10.0 points


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If the mass of the satellite is 215 kg, whatis the minimum energy required to move thesatellite from this near-Earth orbit to very faraway from the Earth?

Correct answer: 6.73131 109 J.

Explanation:Define the system to be the satellite and

the Earth.

W = E

= Ef Ei= Uf + Kf Ui Ki=GMm

rf+

1

2mv2f

GMmri

12

mv2i

Where GMmrf

and 12

mv2f are zero.

=GMm

R 1

2m

GM

R

=1

2

GMm

R

=(G)(6 1024 kg)(215 kg)

2(6.37 106 m + 50000 m)= 6.73131

109 J

004 (part 1 of 2) 10.0 pointsThe figure below is a graph of the energy of asystem of a planet interacting with a star. Thegravitational potential energy Ug is shown asthe thick curve, and plotted along the verticalaxis are various values of K+ Ug.

r1 r2

ABC

K

+

Ug

r, from star to planet

Suppose that K + Ug of the system is A.Which of the following statements are true?

A. The potential energy of the system decreases as the planet moves from r1 tr2.

B. When the separation between the twbodies is r2, the kinetic energy of thsystem is (AB).

C. The system is a bound system; the planecan never escape.D. The planet will escape.E. When the separation between the tw

bodies is r2, the kinetic energy of thsystem is B C).

F. The kinetic energy of the system igreater when the distance between thstar and planet is r1 than when the distance between the two bodies is r2.

Correct answer: B,C,F.

Explanation:Statement A is false because as we move to

the right from r1 to r2, U increases. Statement B is true because the kinetic energy igiven as the total energy minus the potential energy, and since the total energy is Aand U(r2) = B, the kinetic energy is A BStatement C is true because the total energyA is negative, which indicates a bound sys

tem. Statements D-F can be understood fromthe previous three explanations.

005 (part 2 of 2) 10.0 pointsSuppose instead that K + Ug of the systemis B. Which of the following statements artrue?

A. When the separation between the planeand star is r2, the kinetic energy of thsystem is zero.

B. The planet and star cannot get farthe

apart than r2.C. This is not a bound system; the plane

can escape.D. When the separation between the plane

and star is r2, the potential energy of thsystem is zero.

Correct answer: A,B.

Explanation:


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For statement A, refer to the explanationfor statement B for part 1. Here, the totalenergy is KE = B B = 0, so this is true.Statement B is true because in the region be-yond r2, the kinetic energy would be negative,and so this is not allowed. For statement C,

refer to the explanation for statement C frompart 1. The energy is negative and henceits bound; the planet cannot escape. So thisis false. Statement D is false, because thepotential energy at r2 is B.

006 (part 1 of 3) 10.0 points

Given the potential energy curve to the in-teraction of two atoms as shown below. The

system is in a vibrational state indicated bythe heavy horizontal line. Answer the follow-ing questions:

(a) At r = r1, what is the approximatevalue of the potential energy of the molecule?

1. -0.7 eV

2. -0.4 eV

3. -1.0 eV

4. -0.9 eV

5. -1.6 eV

6. -0.2 eV

7. -1.4 eV

8. -1.2 eV

9. -1.3 eV correct

Explanation:(a) The potential energy is about -1.3 eV.

007 (part 2 of 3) 10.0 points(b) At r = r1, what is the approximat

value of the kinetic energy of the molecule?

1. 1.2 eV

2. 1.0 eV

3. 1.5 eV

4. 1.1 eV correct

5. 0.2 eV

6. 0.4 eV

7. 0.7 eV

8. 1.3 eV

9. 0.9 eV

Explanation:

(b) The kinetic energy is K = E U ansince E = 0.2 eV, so K = +1.1 eV.

008 (part 3 of 3) 10.0 points(c) How much energy must be imparte

to the molecule so that it separates into twatoms?

1. 1.6 eV

2. 1.2 eV

3. 1.0 eV

4. 0.4 eV

5. 1.4 eV

6. 0.7 eV

7. 0.9 eV


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8. 1.3 eV

9. 0.2 eV correct

Explanation:(c) The energy to separate (dissociate) the

molecule is 0.2 eV, i.e., enough energy isadded to allow the atoms to have zero en-ergy at large r.

009 (part 1 of 2) 10.0 pointsThe radius of Mars (from the center of justabove the atmosphere) is 3400 km, and itsmass is 6 1023 kg. An object is launchedstraight up from just above the atmosphere ofMars.

What initial speed is needed so that whenthe object is far from Mars, its final speed is1450 m/s? Use G = 6.7 1011 N m2/kg2.

Correct answer: 5074.4.

Explanation:Knowing that energy is conserved and that

Uf = 0, we can write

Ui + Ki = Uf + Kf Ki = Kf Ui

1

2m v2i =

1

2m v2f

G M m

ri

vi =

v2f +2 G M

ri

= 5074.4 m/s .

010 (part 2 of 2) 10.0 pointsWhat initial speed is needed so that when theobject is far from Mars, its final speed is 0m/s? (This is called the escape speed.)

Correct answer: 4862.82.

Explanation:Going through the same process (except

this time Uf andKf are zero), we have

Ui + Ki = 0

Ki = Ui1

2m v2i =

G M m

ri

vi = 2 G Mri= 4862.82 m/s .

011 10.0 pointsThe escape speed from an asteroid whose radius is 8 km is only 7 m/s. If you throwa rock away from the asteroid at a speed o14 m/s, what will be its final speed? UsG = 6.7 1011 N m2/kg2.


Explanation:First use the escape speed to get the mas

of the asteroid:

vesc =

2 G M

R

M = 12

v2 R

G= 2.92537

1015 kg .

Now, if vi = 14 m/s, then vf is found fromEi = Ef:

Ui + Ki = Uf + Kf

G M mri

+1

2m v2i = 0 +

1

2m v2f

vf =

v2i 2 G M

R

= 12.1244 m/s .

012 10.0 pointsYou stand on a spherical asteroid of uniformdensity whose mass is 1 1016 kg and whosradius is 12 km. These are typical valuefor small asteroids, although some asteroidhave been found to have much lower averagdensity and are thought to be loose agglomerations of shattered rocks.

How fast do you have to throw a rock sothat it never comes back to the asteroid and


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ends up traveling at a speed of 3 m/s when it is

very far away? Use G = 6.7 1011 N m2

kg2.


Explanation:

Ei = EfUi + Ki = Uf + Kf

Ki = Uf Ui + Kf

Where Uf is zero.

1

2mv2i = 0

GMmR

+1

2mv2f

1

2

v2i =GM

R

+1

2

v2f

1

2v2i =

(6.7 1011 N m2

kg2)(1 1016 kg)

(12000 m)

+1

2(3 m/s)2

= 55.8333J

kg+ 4.5

J

kg

vi = 10.9848 m/s

013 10.0 pointsA force acting on a particle has the potentialenergy function U(x), shown by the graph.The particle is moving in one dimension underthe influence of this force and has kineticenergy 1.0 Joule when it is at position x1.

x0 x1 x2 x3

1

0

1

Position

PotentialEnergy(J)

Potential Energy vs Position

Which of the following is a correct statement about the motion of the particle?

1. It oscillates with maximum position xand minimum position x0. correct

2. It cannot reach either x0 or x2.

3. It comes to rest at either x0 or x2 anremains at rest.

4. It moves to the right of x3 and does noreturn.

5. It moves to the left of x0 and does noreturn.

Explanation:In this case, the total energy of the particl

is conserved so at any point on the axis,

K(x) + U(x) = K(x1) + U(x1)

= 1.0 J + (

1.0 J) = 0

K(x) = U(x) ,

where K is the kinetic energy of the particleSince kinetic energy 0 ,

K(x) = U(x) 0U(x) 0 J ,

so the particle oscillates between position xand x2.


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014 (part 1 of 2) 10.0 pointsA particle moving along the x-axis has a po-tential energy U(x), as shown in the accom-panying graph.

Scale factors: a = 4.3 m, b = 44 J.

0 a 2 a 3 a 4 a 5 a 6 a

1 b

2 b

3 b

4 b

5 b

U

[J]

x [m]

C

A

B

What is the force exerted on the particlewhen x = 3 a (point A, the lowest point onthe curve)?

Correct answer: 0 N.

Explanation:From the definition of Potential Energy, we

know

F = d Udx

.

In other words, the force on a particle at any

particular point is given by the negative of theslope of the potential energy function at thatpoint.

At x = 3 a, the potential energy has a min-imum; the slope of the function is zero at thatpoint. Therefore there are no forces on a par-ticle at that point. (x = 3 a is an equilibriumpoint.)

015 (part 2 of 2) 10.0 pointsIf the particle has a mass of 7.9 kg and isreleased from rest at x = 5 a, what is theparticles velocity when it reaches x = a?


Explanation:Using the Energy Principle:

Uinit + Kinit = Ufinal + Kfinal

UB + 0 = UC +1

2m v2

Solving for v gives

v =

2

m(UB UC)

= 2

m(3 b 2 b)

=

2

mb

=

2

(7.9 kg)(44 J)

= 3.33755 m/s .

016 10.0 pointsA rock far outside our Solar System is initiallymoving very slowly relative to the Sun, in thplane of Jupiters orbit around the Sun. Throck falls toward the Sun, but on its way tothe Sun it collides with Jupiter. What is thcorrect algebraic expression for the kinetienergy of the rock, Krock immediately beforit hits the surface of Jupiter?

The symbols are defined as follows. Thmasses of Jupiter, of the Sun and the rock ardenoted respectively by M1, M2 and m. Thradii of Jupiter and the Sun are R1 and Rrespectively. The distance between Jupiteand the Sun is d, where dR2R1.

1. Krock =GmM1

R12

GmM2R2

2

2. Krock =GmM1

R12

+GmM2

d2

3. Krock =GmM1

R1+

GmM2d

correct

4. Krock =

GmM1

R22 +

GmM2

R12

5. Krock =GM1M2

d

6. Krock =GmM1

R2+

GmM2R1

7. Krock =GmM1

R1 GmM2

R2

8. Krock =GmM1

R1+

GmM2R2


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9. Krock =GM1M2

d2

10. Krock =GmM1

R12

+GmM2

R22

Explanation:From the energy principle, we have

K + U = 0

Initially the rock moves very slowly and isfar away from both Jupiter and the Sun, soKi 0 and Ui 0. Thus, the energy principlesimplifies to

Kf = Uf1 Uf2

But, we know that

Uf1 = GmM1

R1

Uf2 = GmM2

d

Hence, the final kinetic energy of the rock is

Krock = Kf =GmM1

R1+

GmM2d

017 10.0 pointsFor a satellite and Earth system, which of thefollowing energy graphs of the satellites ki-netic, potential, and total energies as a func-tion of time would be impossible? Assumethat the satellite begins at an altitude of afew Earth radii. Hint: remember that thegravitational potential energy is always nega-tive.

1. G

2. B

3. F correct

4. C

5. D


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6. E

7. A

Explanation:

This graph is an escaping trajectory sincethe total energy is positive. However, in thatcase the potential energy should be decreasingtowards zero, yet it is constant in this graph.Hence, the graph (F) represents the correctanswer.

018 (part 1 of 2) 10.0 pointsA satellite is in a circular orbit a distancer = 3R from the Earths center and has totalenergy E1. What is the fuel energy required

to move it to a circular orbit of radius r = 5R?(Caution: E1 < 0)

1.4

5E1

2. 25

E1 correct

3.2

5E1

4. E1

5. E1

6. 35

E1

7.1

5E1

8.3

5E1

9. 15

E1

10. 45

E1

Explanation:For a circular orbit at radius r, the potential

energy is given by

U = GMmr

The centripetal force is given by

mv2

r=

GMm

r2

From this relation, the kinetic energy is givenby

K = 12

mv2 = GMm2r

= U2

So, the total energy is

E = K+ U = U2

+ U =U

2

E = GMm2r

For the present problem

E1 = GMm6R

E2 = GMm10R

=6

10E1

E2 E1 =

6

10 1

= 410

E1

E2 E1 = 25

E1

019 (part 2 of 2) 10.0 pointsFind the kinetic energy of the satellite in thfinal orbit, where r=5R.

1.3

5E1

2. E1

3.4

5E1

4. 15

E1

5. 45

E1

6. 25

E1

7.2

5E1

8. E1


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9. 15

E1

10. 35

E1 correct

Explanation:

Setting r = 5R

K =GMm

2r=

GMm

10R

K = 610

GMm

6R

= 3

5E1

020 10.0 pointsGiven: G = 6.67259 1011 N m2/kg2

Voyagers 1 and 2 surveyed the surface of

Jupiters moon Io and photographed an activevolcano spewing liquid sulfur to heights of44.4 km above the surface of Io. Ios mass is8.9 1022 kg, and its radius is 1820 km.

If the liquid sulfur reaches 44.4 km, calcu-late the speed with which the liquid sulfur leftthe volcano. Assume the top of the volcano isvery near the surface of Io.


Explanation:The gravitational potential energy at the

moon Ios surface and hence the top of thevolcano is given by

U1 = m GR

M

= m(6.67259 1011 N m2/kg2)

(1820 km)

(8.9 1022 kg)= m (3.26297 106 kg m/s2) .

The gravitational potential energy at the topof the liquid sulfurs trajectory is given by

U2 = mG

R + hM

= m(6.67259 1011 N m2/kg2)

(1820 km) + (44.4 km)

(8.9 1022 kg)= m (3.18526 106 kg m/s2) .

The difference in potential energy is

U=U1 U2= m [(3.26297 106 kg m/s2) (3.18526 106 kg m/s2)]

= m (77706.4 kg m/s2) .

The potential energy of the liquid sulfur athe maximum height it reaches is equal to itinitial kinetic energy as it leaves the surfaceTherefore, we have

K = Um v2

2= m G M

1

R 1

R + h

= m (77706.4 kg m/s2) .

From which it follows that

v =

2 U

=

2 (77706.4 kg m/s2)

= 394.224 m/s .

021 (part 1 of 2) 10.0 pointsWhich of the following choices correspondto a system of two electrons that start oufar apart, moving toward each other (that istheir initial velocities are nonzero and theare heading straight at each other)? Not

that the horizontal and vertical axes in eachplot are the separation between the particleand energy, respectively.

K+ U

U

K(I)

r

K+ U

U

K(II)

r

K+ U

U

K(III)

r

K+ UU

K

(IV)

r


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goldhagen (dfg358) HW-13 florin (57850) 1

K+ U

U

K(V)

rK+ U

U

K

(VI)

r

1. Figure I correct

2. Figure II

3. Figure VI

4. Figure V

5. Figure III

6. Figure IV

Explanation:When the two electrons are very far away

their potential energy is 0, and since theyhave nonzero initial velocities, this means thatthey are unbounded and thus have an overallpositive energy at r = , which is also equalto the kinetic energy at that location. As

the electrons get closer, due to their Coulombrepulsion their kinetic energies drop to 0 whilethe potential energy rises. Thus the correctanswer is Figure (I).

022 (part 2 of 2) 10.0 pointsWhich of the diagrams corresponds to a sys-tem of a proton and an electron that start outfar apart, moving toward each other (that is,their initial velocities are nonzero and theyare heading straight at each other)?

1. Figure I

2. Figure VI

3. Figure II correct

4. Figure V

5. Figure IV

6. Figure III

Explanation:When the two particles are very far away

their potential energy is 0, and since they havnonzero initial velocities, this means that theare unbounded and thus have an overall positive energy at r = , which is also equal tthe kinetic energy at that location. As thelectron and proton get closer, due to theiCoulomb attraction their kinetic energies increase while the negative potential energy decreases even further. Thus the correct answeis Figure (II).

023 10.0 pointsFour protons, each with mass M and charg+e, are initially held at the corners of a squarthat is d on a side.

1 2

3 4d

d

They are then released from rest. What ithe speed of each proton when the protons arvery far apart?

1. v =

q240

1

M d

2 +

2

2

correct

2. v = q2

40

1

Md

3. v =

q240

1

M d

2

2

4. v =

q2

40

1

Md

2

5. v =

q240

1

M d

2

4


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6. v =

q2

40

1

M d

4 +

2

7. v =

q240

1

Md

1 +

2

4

Explanation:Since none of the protons are moving, theyinitially have zero kinetic energy and nonzeropotential energy, and when they are veryfar apart they have 0 potential energy andnonzero kinetic energy. Now for the initialstate, we must include every pairwise interac-tion for the potential, but we must be carefulnot to double-count. The resulting potentialwould be given by the following interactions:1 & 2, 1 & 3, 1 & 4, 2 & 4, 3 & 4, and 2 &

3. Also, because of the symmetry of the prob-lem, each proton will have the same kineticenergy at any given time. We conserve energyand get:

Ef =Ei

KEf + Uf =KEi + Ui

4 KEproton + 0 =0 + U12 + U13+ U14 + U24

+ U34 + U23

4 12

M v2 =1

40

q2

d+

1

40

q2

d

+1

40

q22d

+1

40

q2

d

+1

40

q2

d+

1

40

q22d

2 M v2 = q2

40

4

d+

22d

v =1

2M

q2

404

d +22d

v =

q240

1

Md

2 +

2

2

024 10.0 pointsIn a star, a secondary fusion process in-volves helium-3 and helium-4 fusing togetherinto beryllium-7. Helium-3 has a mass of

3.01603 u, helium-4 has a mass of 4.0026 uand beryllium-7 has a mass of 7.01693 u. Thatomic mass unit u is 1.66054 1027 kgEach one of these fusion reactions will converrest mass energy into kinetic energy. If yowant to have a total of E = 85 J of energy

that has been converted from rest mass energy, how many of these reactions must takplace? Keep six significant figures throughouthis problem, and use c = 2.99792 108 m/s

Correct answer: 3.35028 1014.Explanation:

The rest mass energy released by one othese reactions is

(mHe3 + mHe4 mBe) c2

= 2.5371 101

In order to get the number of reactions tproduce energy E, we divide E by the energper reaction that is found above:

E

2.5371 1013 J =85 J

2.5371 1013 J= 3.35028 1014 .

025 10.0 pointsTwo protons are hurled straight at each othereach with a kinetic energy of 0.45 MeV, wher1 mega electron volt is equal to 1.6 1013 JWhat is the separation from the protons fromeach other when they momentarily come to stop (a turning point)? Use

1

40= 9 109 N m2/C2,

e = 1.6

1019 C.

Correct answer: 1.6 1015 m.Explanation:

Convert the kinetic energy of a proton tojoules to obtain KEi,proton = 7.2 1014 JDefine the system to be the two protons, so

KEi,total = 2(7.2 1014 J)= 1.44 1013 J .


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Assume that they are initially very farapart, so Ui 0. At the final state, theirspeeds are zero at the turning point, which isthe point of closest approach. The work doneon the system is zero and their rest energiesare constant. Thus, the total kinetic energy

is completely converted to the final electricpotential energy.

KEi,total = Uf

Ki =1

40

q1 q2r

Using q1 = q2 = 1.6 1019 C, we obtain

r = 1.6 1015 m .

026 10.0 pointsConsider the decay of a hypothetical particleX, which has a rest mass of 418 Mev/c2, into2 pions,i.e.

X + + Both the pions have the same rest mass,139 Mev/c2.

In the X rest frame, i.e. in the referenceframe where X is at rest, the decaying twopions are moving in opposite directions withthe same speed and the and same energy.

Determine the speed of+ in the X rest frame.Note: Calculate the answer in units ofc.

Hint: In the X rest frame, the total energy

of+ is given bymXc

2

2.

Correct answer: 0.747 c.

Explanation:

let : mX = 418 Mev/c2 and

m = 139 Mev/c2 .

From the energy principle: Ef = Ei =

mXc2. In the rest frame of X, the two pions

have the same momentum p and the sameenergy E. So 2 E = mXc

2. Since

E = m c2 and

=1

1vc

2,

So

2 E = mXc2

2 m c2 = mXc

2

2 m1

1 vc 2

= mX

11

vc

2 = mX2 m1

vc

2=

2 mmxv

c

2= 1

2 mmx

2

v

c

= 1 2 m

mx2

v

c=

1

2 (139 Mev/c2)

418 Mev/c2

v

c= 0.747

v = 0.747 c.

027 (part 1 of 3) 10.0 pointsA 53 kg skier is at the top of a slope, as in th

figure. At the initial point A, the skier is 7.0m vertically above the final point B.

The acceleration of gravity is 9.81 m/s2 .

7.03 m

a) Find the difference in gravitational po

tential energy associated with the skier at thpoints A and B if the zero level for gravitational potential energy is at point B.

Correct answer: 3655.11 J.

Explanation:Basic Concept:

Ug = mgh


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Given:

m = 53 kg

hA = 7.03 m

hB = 0 m

g = 9.81 m/s2

Solution:

UA = mghA= (53 kg)(9.81 m/s2)(7.03 m)

= 3655.11 J

UB = mghB= (53 kg)(9.81 m/s2)(0 m)

= 0 J

Ug = UA

UB

= 3655.11 J 0 J= 3655.11 J

028 (part 2 of 3) 10.0 pointsb) Find the difference in potential energy ifthe zero level is at point A.


Explanation:

Given:

hA = 0 m

hB = 7.03 m

Solution:

UA = mghA= (53 kg)(9.81 m/s2)(0 m)

= 0 J

UB = mghB

= (53 kg)(9.81 m/s2)(7.03 m)= 3655.11 J

Ug = UA UB= 0 J (3655.11 J)= 3655.11 J

029 (part 3 of 3) 10.0 pointsc) Find the difference in potential energy if

the zero level is midway down the slope, at aheight of 3.52 m.


Explanation:Given:

hA = 3.52 mhB = 3.52 m

Solution:

UA = mghA= (53 kg)(9.81 m/s2)(3.515 m)

= 1827.55 J

UB = mghB= (53 kg)(9.81 m/s2)(3.515 m)=

1827.55 J

Ug = UA UB= 1827.55 J (1827.55 J)= 3655.11 J

030 10.0 points

A small mass is released from rest at a verygreat distance from a much larger stationarmass.

Which of the following graphs best repre

sents the gravitational potential energy U othe system of masses as a function of t?

1.

tU

correct

2.

t

U

3.

tU

4.t

U


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5.

tU

6. t

U

7. None of these graphs is correct.

Explanation:Since one of the masses is much larger, the

larger mass will be almost at rest while thesmaller mass will be the one that is accel-erating significantly. Potential energy goesas

U(r) = GM m

r .

Therefore, it rapidly becomes more negativeas r becomes smaller. Moreover, r decreasesat a more and more rapid rate (as the massaccelerates while falling). Hence, the poten-tial energy of the system decreases faster andfaster with time.

A plot of this time functionality U(t) istU

Note: The gravitational potential energy asa function ofr (distance between the masses)is

U(r) = G M mr

.

rU

Date post:	03-Apr-2018
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HW 13 Solutions

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