Ch 8, Potential Energy and Conservation of Energy

Ch8; 9, 11, 13, 21, 26, 31, 52, 57Homework:## Questions 3, 7, 8, 9, 10, 13

Additional problems from Serway's fourth edition

**(4 ed) 8.2 A 3.0-kg block starts at a height h = 60 cm = 0.60 m on a
plane that has an inclination angle of 30 ^{o} as in Figure P8.20.
Upon reaching the bottom, the block slides along a horizontal surface. If
the coefficient of friction on both surfaces is
= 0.20, how far does the block slide on the horizontal surface before coming
to rest?**

**( Hint: Divide the path into two straight-line parts.)**

**(a) the change in its kinetic energy,**

**(b) the change in its potential energy, and**

Conceptual Questions

**Q8.3 A bowling ball is suspended from the ceiling of lecture
hall by a strong cord. The bowling ball is drawn away from its equilibrium position
and released from rest at the top of a student's nose as shown in Figure Q8.3.
If the student remains stationary, explain why she will not be struck by the
ball on its return swing. Would the student be safe if she pushed the ball as
she released?**

As the bowling ball swings back and forth energy is conserved.
Energy changes from gravitational potential energy to kinetic energy and back
again and back and forth. But the **total energy** remains **constant**.
When the student releases the bowling ball **from rest** it has **no kinetic
energy** so its total energy is equal to the gravitational potential energy
at that point. It swings out, stops (meaning its KE is again zero) and then
starts back. At the bottom of its swing it has minimum gravitational potential
energy and maximum kinetic energy. As it swings on -- toward the student --
it is slowing down or loosing kinetic energy as it gains height and gravitational
potential energy. It will completely stop when its gravitational potential energy
equals its total energy -- at that is exactly at the student's chin.

However, if the student pushes the ball that means she gives it
some initial kinetic energy **in addition to** its initial gravitational
potential energy. It will always continue on until its gravitational potential
energy just equals its total energy. This time it will stop at a greater height
since it has a larger amount of total energy. Stopping at a greater height means
it will run into the student's chin!

**Q8.7 If three conservative forces and one nonconservative force
act on a system, how many potential energy terms appear in the equation that
describes this system?**

We can write a potential energy term for each conservative force
so there will be three potential energy terms. We can **not** write a potential
energy term for a nonconservative force. Fricition is a ready example of such
a nonconservative force.

**Q8.8 Consider a ball fixed to one end of a rigid rod whose
other end pivots on a horizontal axis so that the rod can rotate in a vertical
plane. What are the positions of stable and unstable equilibrium?**

**Stable** equilibrium will be with the rod at the bottom of
its arc (or possible arc).

**Unstable** equilibrium will be wit the rod at the **top**
of its arc.

**Q8.9 Is it physicall possible to have a situation where E -
U < 0?**

The total energy E is the **sum** of the potential energy U
and the kinetic energy K,

E = U + K

Therefore, the kinetic energy is the total energy minus the potential energy,

K = E - U

Since K = (1/2) m v^{2}, the kinetic energy is **always
positive**. Therefore, it is **not possible** to have a situation where

E - U < 0

**Q8.10 What would the curve of U versus x look like if a particle
were in a region of neutral equilibrium?**

Neutral equilibrium means **zero net force**. The force is
the **slope** of the U vs x curve. That means the slope is zero so the curve
is a straight, horizontal line.

**Q8.13 If only one external force acts on a particle, does it
necessarily change the particle's**

**(a) kinetic energy?**

Yes.

**(b) velocity?**

Yes (just think of Newton's Second Law, F = m a).

**8.9 A single conservative force acting on a particle varies as F = ( - A
x + B x ^{2}) i N, where A and B are constants and x is in meters.**

**(a) Calculate the potential energy associated with this force, taking U
= 0 at x = 0.**

(b) Find the change in potential energy and change in kinetic energy as the particle moves from x = 2.0 m to x = 3.0 m.U(2) = ( ^{1}/_{2}) A (2)^{2}- (^{1}/_{3}) B (2)^{3}= 2 A - (^{8}/_{3}) BU(3) = (

^{1}/_{2}) A (3)^{2}- (^{1}/_{3}) B (3)^{3}= (^{9}/_{2}) A - 9 BU = U

_{f}- U_{i}= U(3) - U(2) = [ (^{9}/_{2}) A - 9 B ] - [ 2 A - (^{8}/_{3}) B ]U = [ (

^{9}/_{2}) - 2 ] A - [ 9 - (^{8}/_{3}) ] BU = (

^{5}/_{2}) A - (^{19}/_{3}) BE = K + U = constant

E = 0

E = K + U = 0

K = - U

K = - (

^{5}/_{2}) A + (^{19}/_{3}) B

8.11 A 3.0-kg mass starts from rest and slides a distanceddown a frictionless 30^{o}incline, where it contacts an unstressed spring of negligible mass as in Figure P8.30. The mass slides an additional 0.20 m as it is brought momentarily to rest by compressing the spring (k= 400 N/m). Find the initial separationdbetween mass and spring. In describing theinitial potential energy, what do we want tochooseas thereference position?While there are other good choices, I chose to make the potential energyzerowhen the massstopsafter compressing the spring. That is, I choose the reference point for the potential energy -- the location at which the potential energy is zero -- to bed + 20 cm along the plane.

That means the block

startsat an initial height habovethis reference point. And that height is the "opposite side" of a right triangle so thath = (d + 20 cm) sin 30 ^{o}h = (d + 20 cm) (0.500)

h = 0.5 d + 10 cm

h = 0.5 d + 0.10 m

Initially,

E _{i}= K_{i}+ U_{si}+ U_{gi}E

_{i}= 0 + 0 + m g hE

_{i}= (3.0 kg) (9.8 m/s^{2}) (0.5 d + 0.10 m)E

_{i}= 14.7 d (kg-m/s^{2}) + 2.94 JE

_{f}= K_{f}+ U_{sf}+ U_{gf}E

_{f}= 0 + (^{1}/_{2}) k x^{2}+ 0E

_{f}= (^{1}/_{2}) (400 N/m) (0.20 m)^{2}E

_{f}= 8 JE

_{i}= E_{f}E

_{i}= 14.7 d (kg-m/s^{2}) + 2.94 J = 8 J = E_{f}14.7 d (kg-m/s

^{2}) = 5.1 Jd = 0.34 m

d = 34 cm

8.13 A particle of mass m = 5 kg is released from point A and slides on the frictionless track shown in Figure P8.13. Determine

(a) the particle's speed at poins B and C and

(b) the net work done by the force of gravity in moving the particle from point A to point C.

[[ Fig P8.13 ]]E

_{Tot }= KE + UKE = (1/2) m v

^{2}U = m g h

E

_{Tot, A}= 0 + (5 kg)(9.8 m/s^{2})(5 m) = 245 JE

_{Tot}= 245 J = constantE

_{Tot, B}= (1/2)(5 kg)(v_{B}^{2}) + (5 kg)(9.8 m/s^{2})(3.2 m) = E_{Tot}= 245 J(1/2)(5 kg)(v

_{B}^{2}) + 156.8 J = 245 J(1/2)(5 kg)(v

_{B}^{2}) = 245 J - 156.8 J = 88.2 Jv

_{B}^{2}= (35.38)(J/kg)What about the units? Let's check and see.

v

_{B}^{2}= (35.38)(m^{2}/s^{2})v

_{B}= 5.9 m/sOf course, finding the speed at point C is

exactlythe same,E

_{Tot, C}= (1/2)(5 kg)(v_{C}^{2}) + (5 kg)(9.8 m/s^{2})(2.0 m) = E_{Tot}= 245 J(1/2)(5 kg)(v

_{C}^{2}) + 98 J = 245 J(1/2)(5 kg)(v

_{C}^{2}) = 245 J - 98 J =147 Jv

_{C}^{2}= 147 J / 2.5 kgv

_{C}^{2}= (147/2.5)(J/kg)v

_{C}^{2}= (58.8)(J/kg)

v_{C}= 7.7 m/sW

_{AC}= m g hW

_{AC}= (5 kg) ( 9.8 m/s^{2}) (5.0 m - 2.0 m)W

_{AC}= (5 kg) ( 9.8 m/s^{2}) (3.0 m)

W_{AC}= 147 J

8.21 Two masses are connected by a light string passing over a light frictionless pulley as shown in Figure P8.17. The 5.0-kg mass is released from rest.

Using conservation of energy,

(a) determine the speed of the 3.0-kg mass just as the 5.0-kg mass hits the ground.This is essentially the same as 8.10:K _{i}= 0U

_{i}= m_{1}g h_{1}+ m_{2}g h_{2}U

_{i}= (5.0 kg) (9.8 m/s^{2}) (4.0m) + (3.0 kg) (9.8 m/s^{2}) ( 0 )U

_{i}= 196 JE

_{i}= U_{i}+ K_{i}= 196 JE = constant

E

_{f}= E_{i}E

_{f}= K_{f}+ U_{f}K

_{f}= E_{f}- U_{f}U

_{f}= (5.0 kg) (9.8 m/s^{2}) ( 0 ) + (3.0 kg) (9.8 m/s^{2}) (4.0 m)U

_{f}= 117.6 JK

_{f}= 196 J - 117.6 JK

_{f}= 78.4 JK

_{f}= (^{1}/_{2}) (5.0 kg) v^{2}+ (^{1}/_{2}) (3.0 kg) v^{2}K

_{f}= (^{1}/_{2}) (8.0 kg) v^{2}(

^{1}/_{2}) (8.0 kg) v^{2}= 78.4 Jv

^{2}= 19.6 m^{2}/ s^{2}v = 4.43 m/s

Note that bothmasses have the same speed.

(b) Find the maximum height to which the 3.0-kg rises.At a height of 4.0 m, the 3.0-kg mass is moving upwards with a speed of 4.43 m/s. The 5.0-kg mass has just hit the ground so the cord goes loose and there is no other force on the 3.0-kg mass but the force of gravity. How much higher does it go? We know its acceleration so this is akinematicsproblem from long ago. Perhaps the most efficient way is just to usev ^{2}= v_{i}^{2}+ 2 a (s - s_{i})a = - g = - 9.8 m/s

^{2}s

_{i}= 4.0 mv = 0

v

_{i}= 4.43 m/sv

^{2}= v_{i}^{2}+ 2 a (s - s_{i})0 = (4.43 m/s)

^{2}+ 2 ( - 9.8 m/s^{2}) ( s - 4.0 m)0 = 19.6 m

^{2}/s^{2}- (19.6 m/s^{2}) ( s - 4.0 m)(19.6 m/s

^{2}) ( s - 4.0 m) = 19.6 m^{2}/s^{2}s - 4.0 m = 1.0 m

s = 5.0 m

Of course, we could also solve this part of the problem using Conservation of Energy. At a height of 4.0 m, the 3.0-kg mass is moving upwards with a speed of 4.43 m/s. The 5.0-kg mass has just hit the ground so the cord goes loose and there is no other force on the 3.0-kg mass but the force of gravity. How much higher does it go?

For this part of the problem, the "initial conditions" of the 3.0-kg mass are just

v _{i}= 4.43 m/sK

_{i}= (^{1}/_{2}) m v_{i}^{2}= (0.5)(3.0 kg)(4.43^{m}/_{s})^{2}= 29.4 JU

_{i}= m g y_{i}= (3.0 kg)(9.8 m/s^{2})(4.0 m) = 117.6 JE

_{i}= K_{i}+ U_{i}= 147 JE

_{f}= K_{f}+ U_{f}K

_{f}= 0because the block

stopsat its highest point.U _{f}= E_{f}= E_{i}= 147 JU

_{f}= m g y_{f}= 147 J(3 kg)(9.8 m/s

^{2}) y_{f}= 147 Jy

_{f}= 5.0 mAnd, of course, that is exactly what we found from the "kinetmatics approach". The two methods are essentially identical.

8.26 After its release at the top of the first rise, a roller-coaster car moves freely with negligible friction. The roller coaster showin in Fig P8.26 has a circular loop of radius 20.0 m. The car barely makes it around the loop: At the top of the loop, the riders are upside down and feel weightless.

(a) Find the speed of the roller coaster car at the top of the loop (position 3).

Find the speed of the roller coaster car

(b) at position 1 and

(c) at position 2.

(d) Find the difference in height between positions 1 and 4 if the speed at position 4 is 10.0 m/s.

[[ Fig P8.26]]Finding the speed at the top of the loop is a centripetal force problem, straight out of Chapter 6 on Circular Motion. Look back at homework problem

6.19.With the pail of water in problem

6.19, just as with the roller coaster car here, the force of gravity supplies the centripetal force.F

_{c}= F_{g}= w = mgF

_{c}= m v^{2}/r = m gv

^{2}= g rv

^{2}= (9.8 m/s^{2}) (20 m)v

^{2}= 196 m^{2}/s^{2}

v = 14 m/s

v_{3}= 14 m/sThis is v

_{3}the speed at position 3 -- the top of the loop.While not asked for in the text, let's find out the minimum height -- call it position 0 -- from which the roller coaster must start so it can get to this position 3 -- the top of the loop. Now, this -- and the rest of the problem -- just makes use of Energy Conservation.

E = KE + 0

E = (1/2) m v

^{2}+ m g hE

_{0}= 0 + m g h_{0}E

_{3}= (1/2) m v_{3}^{2}+ m g h_{3}E

_{0}= m g h_{0}= (1/2) m v_{3}^{2}+ m g h_{3}= E_{3}h

_{0}= (1/2) v_{3}^{2}/g + h_{3}h

_{0}= (1/2)(14 m/s)^{2}/(9.8 m/s^{2}) + 40 mRemember, h

_{3}is theheightat the top of the loop so that is h_{3}= 2 r = 2 (20 m) = 40 m.

h_{0}= 50 mNow, back to the details asked for in the textbook. What is the speed at position 1, the bottom of the loop? Again, we will use Conservation of Energy.

E = constant

E

_{1}= E_{3}E

_{1}= (1/2) m v_{1}^{2}+ m g h_{1}E

_{3}= (1/2) m v_{3}^{2}+ m g h_{3}

(1/2) m v

_{1}^{2}+ m g h_{1}= (1/2) m v_{3}^{2}+ m g h_{3}

(1/2) v

_{1}^{2}+ g h_{1}= (1/2) v_{3}^{2}+ g h3(1/2) v

_{1}^{2}= (1/2) v_{3}^{2}+ g h_{3}- g h1(1/2) v

_{1}^{2}= (1/2) v_{3}^{2}+ g (h_{3}- h_{1})v

_{1}^{2}= v_{3}^{2}+ 2 g (h_{3}- h_{1})v

_{1}^{2}= (14 m/s)^{2}+ 2 (9.8 m/s^{2})(40 m)v

_{1}^{2}= 980 m^{2}/s^{2}

v_{1}= 31.3 m/sFinding the speed at position 2 is

exactlythe same;E = constant

E

_{2}= E_{3}E

_{2}= (1/2) m v_{2}^{2}+ m g h_{1}E

_{2}= (1/2) m v_{3}^{2}+ m g h_{3}

(1/2) m v

_{2}^{2}+ m g h_{2}= (1/2) m v_{3}^{2}+ m g h_{3}

(1/2) v

_{2}^{2}+ g h_{2}= (1/2) v_{3}^{2}+ g h_{3}(1/2) v

_{2}^{2}= (1/2) v_{3}^{2}+ g h_{3}- g h_{2}(1/2) v

_{2}^{2}= (1/2) v_{3}^{2}+ g (h_{3}- h_{2})v

_{2}^{2}= v_{3}^{2}+ 2 g (h_{3}- h_{2})v

_{2}^{2}= (14 m/s)^{2}+ 2 (9.8 m/s^{2})(20 m)v

_{2}^{2}= 588 m^{2}/s^{2}

v_{2}= 24.3 m/s

8.31 The coefficient of friction between the 3.0-kg mass and surface in Figure P8.31 is 0.40. The system starts from rest. What is the speed of the 5.0-kg mass when it has fallen 1.5 m?E _{i}= K_{i}+ U_{i}E

_{i}= 0 + m_{2}g h_{2}E

_{i}= 0 + (5.0 kg) (9.8 m/s^{2}) (1.5 m)E

_{i}= 73.5 JThis amount of energy goes into the

kinetic energyof the system -- the kinetic energy ofbothmasses --andthe heat associated with the work due to friction.K _{f}= (^{1}/_{2}) m_{1}v^{2}+ (^{1}/_{2}) m_{2}v^{2}= (^{1}/_{2}) (m_{1}+ m_{2}) v^{2}K

_{f}= (^{1}/_{2}) (8 kg) v^{2}= (4 kg) v^{2}U

_{f}= 0E

_{f}= E_{i}+ W_{f}W

_{f}= - F_{f}sF

_{f}= F_{n}F

_{n}= m_{1}gF

_{n}= (3.0 kg) (9.8 m/s^{2})F

_{f}= (0.40) (29.4 N) = 11.8 NW

_{f}= - (11.8 N) (1.5 m)W

_{f}= - 17.6 JE

_{f}= 73.5 J - 17.8 J = 55.7 JE

_{f}= K_{f}= (4 kg) v^{2}= 55.7 Jv

^{2}= 13.9 m^{2}/s^{2}v = 3.7 m/s

8.46 A hollow pipe has one or two weights attache to its inner surface as shown in Figure P8.46. Characterize each configuration as being stable, unstable, or neutral equilibrium and explain each of your choices.(a)isstablebecause a small movement willraisethe center of mass and increase the potential energy of the system.

(b)is inneutralequilibrium because the height of the center of mass will be unchanged by a small movement and, so, the potential energy of the system will be unchanged.

(c)isunstablebecause a small change willlowerthe center of mass and, thus,decreasethe potential energy of the system.

8.52 A 200-g particle is released from rest at point A along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Figure P8.52). Calculate

(a) its gravitational potential energy at point A relative to point B,

(b) its kinetic energy at point B,

(c) its speed at point B, and

(d) its kinetic energy and potential energy at point C.(a)At pointA, the height ish _{A }= Rso the potential energy is

U _{A}= m g h_{A}= m g R

(b)At point B, the height is zero,h _{B}= 0so the potential energy is zero. This means the total energy is all

kinetic energy,U _{B}= 0E

_{B}= K_{B}+ U_{B}= m g R = E_{A}K

_{B}= m g R

(c)We know kinetic energy is given byK _{B}= (^{1}/_{2}) m v_{B}^{2}= m g Rv

_{B}^{2}= 2 g Rv

_{B}= SQRT(2 g R)

(d)At point C, the vertical distance ish _{C}= (^{2}/_{3}) RThat means

U _{C}= m g h_{C}= (^{2}/_{3}) m g RSince the bowl is friction

less, we know energy is conserved. This means the energy at point C is the same as the initial energy at point A;

E _{C}= K_{C}+ U_{C}= E_{A}

K _{C}= E_{A}- U_{C}

K _{C}= m g R - (^{2}/_{3}) m g R

K _{C}= (^{1}/_{3}) m g R

Solutions to the additional problems from Serway's fourth edition.

(4 ed) 8.1 A 5.0-kg mass is attached to a light cord that passes over a massless, frictionless pulley. The other end of the cord is attached to a 3.5-kg mass as shown in Figure P8.*. Use conservation of energy to determine the final speed of the 5.0-kg mass after it has fallen (starting from rest) 2.5 m.K

_{i}= 0U

_{i}= m_{1}g h_{1}+ m_{2}g h_{2}U

_{i}= (5.0 kg) (9.8 m/s^{2}) (2.5 m) + (3.5 kg) (9.8 m/s^{2}) ( 0 )U

_{i}= 122.5 JE

_{i}= U_{i}+ K_{i}= 122.5 JE = constant

E

_{f}= E_{i}E

_{f}= K_{f}+ U_{f}K

_{f}= E_{f}- U_{f}U

_{f}= (5.0 kg) (9.8 m/s^{2}) ( 0 ) + (3.5 kg) (9.8 m/s^{2}) (2.5 m)U

_{f}= 87.25 JK

_{f}= 122.5 J - 87.25 JK

_{f}= 35.25 JK

_{f}= (^{1}/_{2}) (5.0 kg) v^{2}+ (^{1}/_{2}) (3.5 kg) v^{2}K

_{f}= (^{1}/_{2}) (8.5 kg) v^{2}(

^{1}/_{2}) (8.5 kg) v^{2}= 35.25 Jv

^{2}= 8.3 m^{2}/ s^{2}v = 2.9 m/s

Note that

bothmasses have the same speed.

The blockstartswith initial total energy ofE _{i}= U_{i}+ K_{i}E

_{i}= m g h_{i}+ 0E

_{i}= (3 kg) (9.8 m/s2) (0.60 m)E

_{i}= 17.64 J On the way down, that itinital potential energy is changed into kinetic energyandwork done against friction W_{f }. The kinetic energy at the bottom of the incline K_{bot}is found byE _{i}+ W_{f}= E_{bot}E

_{i}= E_{bot}- W_{f}= K_{bot}+ U_{bot}- W_{f}K

_{bot}= E_{i}- U_{bot}+ W_{f}Be (very!) careful of the sign on W

_{f}and where you place it; is this the work donebyfrictiononthe block or is the work doneonthe planebythe block? You can use either -- if it is clear to you which you are using and which sign that interpretation requires. To be consistent with Serway's text, we will take W_{f}to be the work donebyfrictiononthe block which means W_{f}is intrinsicallynegative.W _{f}= - F_{f}sFrom the diagram, we find

sin 30 ^{o}= 0.500 =^{opp}/_{hyp}s = hyp =

^{opp}/_{0.5}=^{0.60 m}/_{0.5}s = 1.2 m

F

_{f}= F_{n }Also, from the diagram, we find that

F _{n }= m g cos 30^{o}= (3 kg) (9.8 m/s^{2}) (0.866)F

_{n}= 25.5 NF

_{f}= (0.20) (25.5 N) = 5.1 NW

_{f}= - F_{f}sW

_{f}= - (5.1 N) (1.2 m)W

_{f}= - 6.11 JK

_{bot}= E_{i}- U_{bot}+ W_{f}K

_{bot}= 17.64 J - 0 - 6.11 JK

_{bot}= 11.53 JThis is the

kinetic energyat the bottom. But, since that is our reference position for the potential energy, the potential energy at the bottom is zero. So this is thetotal energy-- the total, large-scale, mechanical energy -- of the system. 6.11 J of initial energy has gone into heat as friction did work on the block and plane. Now, on the horizontal surface, the normal force F_{n}and the friction force F_{f}have changed;F _{n}= m g = (3.0 kg) (9.8 m/s^{2})F

_{n}= 29.4 NF

_{f}= F_{n}= (0.20) (29.4 N) = 5.9 NW

_{net}= W_{f}= - F_{f}s_{hor}W

_{net}= K = K_{final}- K_{bot}= 0 - K_{bot}= - K_{bot}- F

_{f}s_{hor}= - K_{bot}(5.9 N) s

_{hor}= 11.5 Js

_{hor}= 1.96 m

(4 ed) 8.3 A toy consists of a piece of plastic attached to a spring as in Figure P8.*. The spring is compressed 2.0 cm and the toy is released. If the mass of the toy is 100 g and it rises to a maximum height of 60 cm, estimate the force constant of the spring.

Initially, there is only spring potential energy stored in the spring,E _{i}= U_{si}= (^{1}/_{2}) k x^{2}= (^{1}/_{2}) k (0.02 m)^{2}Later, at the top of its "jump", when its velocity is momentarily zero, there is only gravitational potential energy,

E _{f}= U_{gf}= m g h_{f}= (0.1 kg) (9.8 m/s^{2}) (0.60 m) = 0.59 JSince we expect energy to be conserved, this means

E _{i}= E_{f}(

^{1}/_{2}) k (0.02 m)^{2}= 0.59 Jk = 2 940 N / m

(4 ed) 8.4 A particle of massmstarts from rest and slides down a frictionless track as in Figure P8.*. It leaves the track horizontally, striking the ground as indicated in the sketch. Determineh. What is the horizontal velocity of the ball as it leaves the track? How long does it take (how muchtimeis required) for the ball tofall1.25 m?y = y _{o}+ v_{yo}t + (^{1}/_{2}) a_{y}t^{2}- 1.25 m = 0 + 0 + (

^{1}/_{2}) ( - 9.8 m/s^{2}) t^{2}t

^{2}= 0.255 s^{2}t = 0.505 s

During this time, the ball travels 1.0 m so its horizontal speed must be

v = v _{x}= 1.0 m /0.505 sv = 1.98 m/s

That means its kinetic energy at the bottom of the track is

K = ( ^{1}/_{2}) m v^{2}= (^{1}/_{2}) m (1.98 m/s)^{2}At the bottom of the track, the gravitational potential energy is

zero; that is, we havechoosenthebottomof the track as our reference point. That means the total energy is just the kinetic energy,E _{bot}= K_{bot}+ U_{bot}= K_{bot}+ 0 = K_{bot}= (^{1}/_{2}) m (1.98 m/s)^{2}At the

topof the track, when the ball wasat rest, the kinetic energy was zeroE _{top}= K_{top}+ U_{top}= 0 + U_{top}= m g hEnergy conservation means

E _{top}= E_{bot}m g h = (

^{1}/_{2}) m (1.98 m/s)^{2}h = 1.96 m

(4 ed) 8.4 A 4.0-kg particle moves along the x-axis under the influence of a single conservative force. If the work done on the particle is 80.0 J as it moves from the point x = 2.0 m to x = 5.0 m, find

(a) the change in its kinetic energy,

(b) the change in its potential energy, and

(c) it speed at x = 5.0 m if it starts from rest at x = 2.0 mW = W _{net}= 80 JW = W

_{net}= KK = 80 J

U = - W

U = - K

U = - 80 J

K(x=2) = 0

K(x=5) = K

_{f}= 80 JK(x=5) = (

^{1}/_{2}) m v^{2}= 80 J(

^{1}/_{2}) (4 kg) v^{2}= 80 Jv

^{2}= 40 m^{2}/ s^{2}v = 6.32 m / s

(c) Doug Davis, 2001; all rights reserved