A.  Z.  ALZAHRANI    

Chapter 8: Conservation of Energy

This chapter actually completes the argument established in the previous chapter and

outlines the standing concepts of energy and conservative rules of total energy. I will

divide the chapter into two main parts: conservative and nonconservative forces and

conservation of energy.

A- Conservative and Nonconservative Forces There are two basic properties of the conservative forces that are related to the work

done by this force. These are:

• If the work done by the force is independent of the path, the force is said to

be a conservative force. Otherwise, it is nonconservative.

• The work done by a conservative force in any closed path is zero. The closed

path means a zero displacement.

As examples of conservative forces, the gravitational and spring forces do. On the

other side, friction force gives an example of nonconservative forces.

B- Conservation Law of Mechanical Energy The law of conservation of energy states that: the total mechanical energy of a

system remains unchanged (constant) in any isolated system of objects that

interact only through conservative forces. In other words, it states that: the energy

may neither be created nor destroyed, but transferred from a form of energy to

another.

The total mechanical energy, E, is defined as:

E = K + U

Where K is the total kinetic energy and U is the total potential energy of the system.

A.  Z.  ALZAHRANI    

The conservation law of energy is , therefore,

Ei = Ef

Where Ei and Ef are the initial and final energies, respectively. That implies to

Ki + Ui = Kf + Uf

The potential energy is defined (from chapter 7) as:

U= Wg = F. d = mg . d=mg d cosθ

Whereas the kinetic energy is defined as:

K = ½ m v2

In the following diagram, kinetic and potential energies are calculated for a car

starting its motion from rest at the top of a cliff. The total energy of the car is

conserved at each point till it reaches the bottom of the cliff.

A.  Z.  ALZAHRANI    

Examples:

1. Using the work-energy theorem, show that W= -∆U.

Solution The work-energy theorem states that

W = ∆K

However, the law of conservation of energy is:

Ki + Ui = Kf + Uf

It can be rewritten as

Kf - Ki = Ui - Uf = -(Uf – Ui)

Or

∆K = - ∆U

Therefore,

W = - ∆U

2. Find the height from which you would have to drop a ball so that it

would have a speed of 9.0 m/s just before it hits the ground.

Solution The initial energy of the system is

Ei = m gh

The final energy of the system is

Ef = ½ mv2

Therefore,

½ mv2= m gh

A.  Z.  ALZAHRANI    

Hence,

h= v2/2g = 9.02/19.6 = 4.1 m

3. A ball is thrown vertically upwards and reaches a maximum height of

19.6 m. Calculate its initial speed.

Solution The initial energy of the system is

Ei = ½ mv2

The final energy of the system is

Ef = m gh

Therefore,

½ mv2= m gh

v = [2 gh]0.5=[2 × 9.8 × 19.6]0.5=19.6 m/s

3. A block having a mass of 4.0 kg is given an initial velocity vA =2.0 m/s

to the right and collides with a spring of force constant k= 64.0 N/m.

Calculate the maximum compression of the spring after the collision,

assuming the surface is frictionless.

Solution

The initial energy, Ei, is

Ei = Ki + Ui

8.5 Work Done by Nonconservative Forces 229

Figure 8.12 A block sliding on a smooth, horizontal surface col-lides with a light spring. (a) Initially the mechanical energy is all ki-netic energy. (b) The mechanical energy is the sum of the kinetic energy of the block and the elastic potential energy in the spring. (c) The energy is entirely potential energy. (d) The energy is trans-formed back to the kinetic energy of the block. The total energy re-mains constant throughout the motion.

Multiflash photograph of a pole vault event. Howmany forms of energy can you identify in this picture?

the maximum compression of the spring, which in this casehappens to be xC . The total mechanical energy of the systemis conserved because no nonconservative forces act on ob-jects within the system.

Because mechanical energy is conserved, the kinetic en-ergy of the block before the collision must equal the maxi-mum potential energy stored in the fully compressed spring:

Note that we have not included Ug terms because no changein vertical position occurred.

(b) Suppose a constant force of kinetic friction acts be-tween the block and the surface, with If the speed!k " 0.50.

0.15 m"

xm " ! mk

vA " ! 0.80 kg50 N/m

(1.2 m/s)

12mvA

2 # 0 " 0 # 12kxm

2

KA # UsA " KC # UsC

EA " EC

of the block at the moment it collides with the spring is 1.2 m/s, what is the maximum compression in the spring?

Solution In this case, mechanical energy is not conservedbecause a frictional force acts on the block. The magnitudeof the frictional force is

Therefore, the change in the block’s mechanical energy dueto friction as the block is displaced from the equilibrium posi-tion of the spring (where we have set our origin) to xB is

Substituting this into Equation 8.15 gives

Solving the quadratic equation for xB gives m andm. The physically meaningful root is

The negative root does not apply to this situation

because the block must be to the right of the origin (positivevalue of x) when it comes to rest. Note that 0.092 m is lessthan the distance obtained in the frictionless case of part (a).This result is what we expect because friction retards the mo-tion of the system.

0.092 m.

xB "xB " $0.25xB " 0.092

25xB

2 # 3.92xB $ 0.576 " 0

12(50)xB

2 $ 12(0.80)(1.2)2 " $3.92xB

%E " Ef $ Ei " (0 # 12kxB

2) $ (12mvA

2 # 0) " $ fkxB

%E " $ fkxB " $3.92xB

fk " !kn " !kmg " 0.50(0.80 kg)(9.80 m/s2) " 3.92 N

vA "

E = – mvA21

2

x = 0

(a)

(b)

(c)

vC = 0

(d)

xm

!

"

#

$

E = – mvB2 + – kxB

212

12

E = – mvD2 = – mvA

212

12

E = – kxm21

2

vA

vB

xB

vD = –vA

A.  Z.  ALZAHRANI    

There is only kinetic energy, therefore

Ei = ½ mv2

The final energy, Ef, is

Ef = Kf + Uf

While we have asked about the maximum compression, that means the

body will momentarily stop. Therefore, at this point, there is only

potential energy due to the spring. Hence,

Ef = ½ kx2

From the conservation law, we have

Ei = Ef

Therefore,

½ mv2 = ½ kx2

That gives

x = (mv2/k)0.5 =(4 × 22/64)0.5 =0.5 m

4. A 4.00 kg particle is freely released from point (A) and slides on the

frictionless. Determine (a) the particle’s speed at points (B) and (C) and

(b) the total work done by the force of gravity in moving the particle from

(A) to (C).

Solution

Problems 241

Figure P8.15

Figure P8.13

Figure P8.11 Problems 11 and 12.

12. A mass m starts from rest and slides a distance d down africtionless incline of angle !. While sliding, it contactsan unstressed spring of negligible mass, as shown in Fig-ure P8.11. The mass slides an additional distance x as itis brought momentarily to rest by compression of thespring (of force constant k). Find the initial separationd between the mass and the spring.

cal spring of constant k " 5 000 N/m and is pusheddownward so that the spring is compressed 0.100 m. Af-ter the block is released, it travels upward and thenleaves the spring. To what maximum height above thepoint of release does it rise?

18. Dave Johnson, the bronze medalist at the 1992 Olympicdecathlon in Barcelona, leaves the ground for his highjump with a vertical velocity component of 6.00 m/s.How far up does his center of gravity move as he makesthe jump?

19. A 0.400-kg ball is thrown straight up into the air andreaches a maximum altitude of 20.0 m. Taking its initialposition as the point of zero potential energy and usingenergy methods, find (a) its initial speed, (b) its totalmechanical energy, and (c) the ratio of its kinetic en-ergy to the potential energy of the ball–Earth systemwhen the ball is at an altitude of 10.0 m.

20. In the dangerous “sport” of bungee-jumping, a daringstudent jumps from a balloon with a specially designed

14. A simple, 2.00-m-long pendulum is released from restwhen the support string is at an angle of 25.0° from thevertical. What is the speed of the suspended mass at thebottom of the swing?

15. A bead slides without friction around a loop-the-loop(Fig. P8.15). If the bead is released from a height h "3.50R, what is its speed at point A? How great is the nor-mal force on it if its mass is 5.00 g?

16. A 120-g mass is attached to the bottom end of an un-stressed spring. The spring is hanging vertically and hasa spring constant of 40.0 N/m. The mass is dropped.(a) What is its maximum speed? (b) How far does itdrop before coming to rest momentarily?

17. A block of mass 0.250 kg is placed on top of a light verti-

13. A particle of mass m " 5.00 kg is released from point !and slides on the frictionless track shown in FigureP8.13. Determine (a) the particle’s speed at points "and # and (b) the net work done by the force of gravityin moving the particle from ! to #.

m = 3.00 kg

d

k = 400 N/m

! = 30.0°!

3.20 m

!

"

#

m

2.00 m

5.00 m

A

R

h

Figure P8.20 Bungee-jumping. (Gamma)

A.  Z.  ALZAHRANI    

(a) The initial energy at point (A), EiA, is

EiA = KiA + UiA

Because the body starts from rest, there is only potential energy

Ei = UiA

The final energy at point (B), EfB, is

EfB = KfB + UfB

At this point we have both kinetic and potential energies. From the

conservation law, we find

EiA = EfB

Or

UiA = KfB + UfB

That means

KfB = UiA - UfB

Therefore,

½ mvB2 = mg (dA -dB)

vB = [2 g (dA -dB)]0.5= [2×9.8×(5.0-3.2)]0.5 =5.95 m/s

The final energy at point (C), EfC, is

EfC = KfC + UfC

However,

EiA = EfC

Or

UiA = KfC + UfC

That means

KfC = UiA - UfC

A.  Z.  ALZAHRANI    

Therefore,

½ mvC2 = mg (dA –dC)

vC = [2 g (dA –dC)]0.5= [2×9.8×(5.0-2.0)]0.5 =7.68 m/s

(b) The work done by the gravity between points A and C is

WAC = - ∆UAC = - (UC – UA)

Therefore

WAC = - mg (dC – dA)= - 2×9.8×(5.0 -2.0) = -58.8 J

5. Two masses are connected by a light string, which is passing over a

light frictionless pulley. The mass m1 is released from rest. Using the law

of conservation of energy, determine the speed of m2 just as m1 hits the

ground.

Solution

The two masses will move under the influence of a similar acceleration.

Therefore their velocities will be the same when the mass m1 hits the

ground. Hence analysing each of them, separately, gives the followings:

242 C H A P T E R 8 Potential Energy and Conservation of Energy

elastic cord attached to his ankles, as shown in FigureP8.20. The unstretched length of the cord is 25.0 m, thestudent weighs 700 N, and the balloon is 36.0 m abovethe surface of a river below. Assuming that Hooke’s lawdescribes the cord, calculate the required force constantif the student is to stop safely 4.00 m above the river.

21. Two masses are connected by a light string passing over alight frictionless pulley, as shown in Figure P8.21. The5.00-kg mass is released from rest. Using the law of con-servation of energy, (a) determine the speed of the 3.00-kg mass just as the 5.00-kg mass hits the ground and (b)find the maximum height to which the 3.00-kg mass rises.

22. Two masses are connected by a light string passing overa light frictionless pulley, as shown in Figure P8.21. Themass m1 (which is greater than m2) is released from rest.Using the law of conservation of energy, (a) determinethe speed of m2 just as m1 hits the ground in terms ofm1, m2, and h, and (b) find the maximum height towhich m2 rises.

cal circular arc (Fig. P8.25). Suppose a performer withmass m and holding the bar steps off an elevated plat-form, starting from rest with the ropes at an angle of !iwith respect to the vertical. Suppose the size of the per-former’s body is small compared with the length !, thatshe does not pump the trapeze to swing higher, and thatair resistance is negligible. (a) Show that when the ropesmake an angle of ! with respect to the vertical, the per-former must exert a force

in order to hang on. (b) Determine the angle !i at whichthe force required to hang on at the bottom of the swingis twice the performer’s weight.

F " mg (3 cos ! # 2 cos !i)

Figure P8.25

Figure P8.21 Problems 21 and 22.

23. A 20.0-kg cannon ball is fired from a cannon with amuzzle speed of 1 000 m/s at an angle of 37.0° with thehorizontal. A second ball is fired at an angle of 90.0°.Use the law of conservation of mechanical energy tofind (a) the maximum height reached by each ball and(b) the total mechanical energy at the maximum heightfor each ball. Let y " 0 at the cannon.

24. A 2.00-kg ball is attached to the bottom end of a lengthof 10-lb (44.5-N) fishing line. The top end of the fishingline is held stationary. The ball is released from restwhile the line is taut and horizontal (! " 90.0°). Atwhat angle ! (measured from the vertical) will the fish-ing line break?

25. The circus apparatus known as the trapeze consists of abar suspended by two parallel ropes, each of length !.The trapeze allows circus performers to swing in a verti-

26. After its release at the top of the first rise, a roller-coaster car moves freely with negligible friction. Theroller coaster shown in Figure P8.26 has a circular loopof radius 20.0 m. The car barely makes it around theloop: At the top of the loop, the riders are upside downand feel weightless. (a) Find the speed of the rollercoaster car at the top of the loop (position 3). Find thespeed of the roller coaster car (b) at position 1 and (c) at position 2. (d) Find the difference in height be-tween positions 1 and 4 if the speed at position 4 is 10.0 m/s.

27. A light rigid rod is 77.0 cm long. Its top end is pivotedon a low-friction horizontal axle. The rod hangs straightdown at rest, with a small massive ball attached to itsbottom end. You strike the ball, suddenly giving it a hor-izontal velocity so that it swings around in a full circle.What minimum speed at the bottom is required tomake the ball go over the top of the circle?

h " 4.00 mm2 " 3.00 kg

m1 " 5.00 kg

"

!

A.  Z.  ALZAHRANI    

For the mass, m1, the initial kinetic and potential energies are

Ki1 = 0

And

Ui1 = -m1 gh

Therefore,

Ei1 = -m1 gh

The final kinetic and potential energies (when it hits the ground) are

Kf1 = ½ m1v2

And

Uf1 = 0

Therefore,

Ef1 = ½ m2v2

For the mass, m2, the initial kinetic and potential energies are

Ki2 = Ui2 = 0

Therefore,

Ei2 = 0

The final kinetic and potential energies are

Kf2 = ½ m2v2

And

Uf2 = - m2 gh

Therefore,

Ef2 = ½ m2v2 - m2 gh

The initial total energy of the system is

Ei = Ei1 + Ei2 = -m1 gh

A.  Z.  ALZAHRANI    

The final total energy of the system is

Ef = Ef1 + Ef2 = ½ m1v2 + ½ m2v2 - m2 gh

It is known from the conservation law of energy that

Ei = Ef

That implies to

½ m1v2 + ½ m2v2 - m2 gh = -m1 gh

Therefore,

½ (m1+ m2) v2= (m2 – m1) gh

v = [2 (m2 - m2)/ (m1 + m2) gh]0.5

That gives

v = [2×9.8×4.0×(5.0-3.0)/(5.0+3.0)]0.5 =4.43 m/s

6. Two bodies are connected by a cord, which passes through a small

pulley. The coefficient of friction between the 3.00-kg block and the

surface is 0.4. Estimate the speed of the 5.00-kg ball when it has fallen

1.50 m?

Solution

For the mass, m1=5.00 kg, the initial energy

Ki1 = Ui1 = 0

Problems 243

Section 8.5 Work Done by Nonconservative Forces28. A 70.0-kg diver steps off a 10.0-m tower and drops

straight down into the water. If he comes to rest 5.00 mbeneath the surface of the water, determine the averageresistance force that the water exerts on the diver.

29. A force Fx , shown as a function of distance in FigureP8.29, acts on a 5.00-kg mass. If the particle starts fromrest at x ! 0 m, determine the speed of the particle at x ! 2.00, 4.00, and 6.00 m.

32. A 2 000-kg car starts from rest and coasts down from thetop of a 5.00-m-long driveway that is sloped at an angleof 20.0° with the horizontal. If an average friction forceof 4 000 N impedes the motion of the car, find thespeed of the car at the bottom of the driveway.

33. A 5.00-kg block is set into motion up an inclined planewith an initial speed of 8.00 m/s (Fig. P8.33). The blockcomes to rest after traveling 3.00 m along the plane,which is inclined at an angle of 30.0° to the horizontal.For this motion determine (a) the change in the block’skinetic energy, (b) the change in the potential energy,and (c) the frictional force exerted on it (assumed to beconstant). (d) What is the coefficient of kinetic friction?

Figure P8.33

Figure P8.31

Figure P8.29

Figure P8.26

34. A boy in a wheelchair (total mass, 47.0 kg) wins a racewith a skateboarder. He has a speed of 1.40 m/s at thecrest of a slope 2.60 m high and 12.4 m long. At the bot-tom of the slope, his speed is 6.20 m/s. If air resistanceand rolling resistance can be modeled as a constant fric-tional force of 41.0 N, find the work he did in pushingforward on his wheels during the downhill ride.

35. A parachutist of mass 50.0 kg jumps out of a balloon ata height of 1 000 m and lands on the ground with aspeed of 5.00 m/s. How much energy was lost to air fric-tion during this jump?

36. An 80.0-kg sky diver jumps out of a balloon at an alti-tude of 1 000 m and opens the parachute at an altitudeof 200.0 m. (a) Assuming that the total retarding force

30. A softball pitcher swings a ball of mass 0.250 kg arounda vertical circular path of radius 60.0 cm before releas-ing it from her hand. The pitcher maintains a compo-nent of force on the ball of constant magnitude 30.0 Nin the direction of motion around the complete path.The speed of the ball at the top of the circle is 15.0 m/s.If the ball is released at the bottom of the circle, what isits speed upon release?

31. The coefficient of friction between the 3.00-kg blockand the surface in Figure P8.31 is 0.400. The systemstarts from rest. What is the speed of the 5.00-kg ballwhen it has fallen 1.50 m?

1

2

34

876543210 x(m)12345

Fx(N)

3.00 kg

5.00 kg

3.00 mvi = 8.00 m/s

30.0°

WEB

A.  Z.  ALZAHRANI    

Therefore,

Ei1 = 0

The final kinetic and potential energies are

Kf1 = ½ m1v2

Uf1 = -m1 gh

Therefore,

Ef1 = ½ m1v2 - m1 gh

For the mass, m2=3.00 kg, we know that the total work is equivalent to

the change in the kinetic energy

W = ∆K = - ∆U

The work done by the friction force only, therefore

W = -µkm2 gd

Hence

∆K= -µkm2 gd

From the conservation law of energy we get

½ m1v2 - m1 gh = -µk m2 gd

Or

v = [2 (m1 - µkm2)/ (m1 + m2) gd]0.5

v = [2 (5.0 – 0.4 × 3.0)/ (5.0+3.0) × 9.8 ×1.5]0.5 =3.73 m/s

A.  Z.  ALZAHRANI    

7. A motorcyclist is trying to jump across the valley by driving

horizontally off the cliff at a speed of 38.0 m/s. Find the speed with which

the cycle strikes the ground on the other side, as shown in the figure.

Solution

The initial energy at the top, Ei, is

Ei = Ki + Ui = ½ mv02 - mg h0

The final energy at the bottom, Ef, is

Ef = Kf + Uf = ½ mvf2 - mg hf

From the conservation law, we find

½ mv02 - mg h0 = ½ mvf

2 - mg hf

Then we get

vf = [v02 – 2 g (h0 - hf)]0.5

vf = [38.02 – 2 × 9.8 (70.0 – 35.0)]0.5 =27.53 m/s

A.  Z.  ALZAHRANI    

8. A person is sitting on a sledge at the top of a 23.7 m tall hill. Determine

their speed when they reach the bottom of the hill.

Solution The initial energy at the top, Ei, is

Ei = Ki + Ui = 0 + mg h

The final energy at the bottom, Ef, is

Ef = Kf + Uf = ½ mv2 + 0

From the conservation law, we find

½ mv2 = mg h

Then we get

v = [2 g h]0.5

v = [2 × 9.8 × 23.7]0.5 = 21.6 m/s