CHAPTER 7

CONSERVATION OF ENERGY

• Conservation of mechanical energy

• Conservation of total energy of a system

• Examples

• Origin of friction

Potential energy functions used in Chapter 7

Gravitational potential energy

The gravitational potential energy of a mass M at a

distance h above the zero of potential energy (usually the

ground is

Elastic potential energy

When a spring is compressed (or stretched) a distance

from its equilibrium (i.e., unstretched) length, the elastic

potential energy of the spring is

where k is the spring constant.

UG = Mgh.

UE = 1

2kℓ2,

ℓ

Conservation of mechanical energy

Consider the motion of an object in one dimension. In the

previous chapter we found that the work done by a force

moving the object from point 1 to point 2 is

where K is the kinetic energy. Also, if the force is

conservative, then

where is the change in potential energy from point 1 to

point 2.

Defining the total mechanical energy as

then

i.e., the total energy remains unchanged. This is a statement

of the conservation of mechanical energy.

W12 = F

1

2∫ dx = ΔK = K2 − K1,

F

1

2∫ dx = −ΔU = −(U2 − U1),

∴ΔK = K2 − K1 = −ΔU = −(U2 − U1),

i.e., K2 + U2 = K1 + U1.

E = K + U,

ΔE = (K2 + U2) − (K1 + U1) = 0,

ΔU

Conservation of mechanical energy provides us with an

alternative way to determine velocities, displacements, etc.,

as illustrated in the following.

An object is fired vertically from the ground with an initial

velocity What is its velocity at a height y above the

ground? What is the maximum height h it reaches?

Let the gravitational potential energy at

ground level The kinetic energy

of the object at ground level is

so the mechanical energy is

Since the gravitational force is conservative, mechanical

energy is conserved, so we put

v!.

y h

K1 =

12

mv!2,

U1 = 0.

E1 = K1 + U1 =

12

mv!2.

E1 = E.

At a distance y above the ground, the kinetic energy is

and the gravitational potential energy is

Since mechanical energy is conserved,

i.e.,

At maximum height, so Also

Using the conservation of mechanical energy

We would get the same results using the kinematic

equations derived in chapter 2.

K2 = 1

2mv2,

U2 = mgy.

K2 + U2 = 1

2mv2 + mgy = E = 1

2mv!

2,

v2 = v!2 − 2gy.

∴v = v!2 − 2gy.

v = 0, K3 = 0. U3 = mgh.

K3 + U3 = mgh = E = 12

mv!2.

∴h =v!

2

2g.

K = 1

2mv2

UG = mgy

See how the individual components of the mechanical

energy vary in time as the object rises. Since

i.e., the decrease in kinetic energy equals the increase

in gravitational potential energy (and vice versa when

an object is dropped).

E = K + U ⇒ constant (in time)ΔE = ΔK + ΔU = 0.∴ΔK = −ΔU,

y

Question 7.1: Two children throw stones from the top of

a building at the same instant with the same initial speed.

One is thrown horizontally, the other is thrown vertically

upward. Which of the following statements best

describes what happens. The stones will strike the

ground …

A: at the same time with equal speeds,

B: at different times with equal speeds,

C: at the same time with different speeds,

D: at different times with different speeds.

h

This involves the conservation of mechanical energy. The

change in gravitational potential energy is

converted to kinetic energy. Therefore, for each stone:

Putting at ground level, we get

so their final speeds

are the same as they started from the same height (h)! Since

the speeds are the same at the ground, we have

where is the time for the stone thrown horizontally and

is the time for the stone thrown vertically. So,

Therefore, the answer is B – different times with equal

speeds (and independent of mass).

(ΔU = −mgh)

Kf + Uf = Ki + Ui.

Uf = 0

12

mvf2 = 1

2mvi

2 + mgh,

i.e., vf2 = vi

2 + 2gh,

vf = vi2 + 2gh,

0 − gt1 = vy − gt2, i.e., t1 = t2 −

vyg

,

t1

t2 t1 < t2.

The following two examples involve two forms of

potential energy; gravitational and elastic. Note that both

are associated with conservative forces and so we can use

the conservation of mechanical energy.

Question 7.2: A 1.0 kg block is placed on top of a vertical

spring with its lower end fixed to the ground. The spring

has an unstretched length of 20.0 cm and spring constant

The spring is then compressed to a length

of 8.0 cm and released so the block is shot vertically

upwards. What height did the block reach?

k = 1200 N/m.

Before [1]

y1 = 0.08 mv1 = 0m = 1.0 kg

After [2]

y2 = ?v2 = 0k = 1200 N/m

We choose the ground to be the zero of gravitational

potential energy. At position [1]

and at position [2]

Using the conservation of mechanical energy

i.e.,

K1 = 0 : UG1 = mgy1 : UE1 =

12

k(Δy)2,

K2 = 0 : UG2 = mgy2 : UE2 = 0.

K2 + UG2 + UE2 = K1 + UG1 + UE1,

mgy2 = mgy1 +

12

k(Δy)2.

∴y2 = y1 +12

kmg

(Δy)2

= (0.08 m) + (1200 N/m)(0.12 m)2

2(1.0 kg)(9.81 m/s2)= 0.96 m.

Question 7.3: A 3.00 kg object is released from rest at a

height of 5.00 m on a curved frictionless ramp, as shown

above. At the foot of the ramp is a spring with a spring

constant The block slides down the ramp

compressing it distance x before coming momentarily to

rest. What is the distance x?

k = 400 N/m.

k = 400 N/m

x

5.00 m k = 400 N/m

x

5.00 m

UG = 0

At the top of the ramp:

At the bottom of the ramp:

Mechanical energy is conserved …

K = 0 : UG = mgh : UE = 0.

∴Etop = mgh = (3.00 kg)(9.81 m/s2)(5.00 m)

= 147.2 J.

K = 0 : UG = 0 : UE = 12

kx2.

∴Ebottom = 12

(400 N/m)x2 = 200x2 J.

∴Ebottom = Etop,

i.e., 200x2 = 147.2 J.∴x = 0.86 m.

Question 7.4: (A conservation of energy problem

involving the potential energy function.) An object of

mass 2.5 kg is confined to move along the x-axis. If the

potential function is

where is in Joules and x is in meters and the

velocity of the object is 2.0 m/s at what is its

velocity at

U(x) = 3x2 − 2x3,

U(x)

x = −1 m,

x = 1 m?

Since a conservative force is involved – how do we know?

– the total mechanical energy is conserved. At

the kinetic energy is

and the potential energy is

So, the total energy at is

Therefore, the kinetic energy at is

x = −1 m

K1 =

12

mv12 = 1

2(2.5 kg)(2.0 m/s)2 = 5.0 J,

U1 = 3(−1)2 − 2(−1)3⎡

⎣⎢⎤⎦⎥ J = 5.0 J.

x = −1 m

Emech = 10.0 J.

x = 1 m

K2 = Emech − U2

= 10.0 J − 3(1)2 − 2(1)3⎡⎣⎢

⎤⎦⎥ J = 9.0 J.

∴v2 =2K2

m= 2(9.0 J)

2.5 kg= 2.68 m/s.

DISCUSSION PROBLEM [7.1]:

Two identical balls roll along similar tracks, the only

difference is that track B has a small depression in it but

the overall distances the balls travel from start to finish are

the same. Since the balls “fall” the same overall heights

(h) conservation of mechanical energy tells us that,

ignoring friction, their speeds are the same at the end

points. But if it’s a race, which one (if either) reaches the

finish line first?

Question 7.5: A pendulum consists of a 2.0 kg mass

hanging vertically from a string of negligible mass with a

length of 3.0 m. The mass is struck horizontally so it has

an initial horizontal velocity of 4.5 m/s. At the point

where the string makes an angle of with the vertical,

what is

(a) the gravitational potential energy of the mass,

(b) the speed of the mass, and

(c) the tension in the string?

(d) What is the angle of the string with the vertical

when the mass is at its greatest height?

30!

The mechanical energy at all points of the swing is

Take the zero of potential energy at , i.e., put

Then

at

at

(a) From the figure we deduce that

E = K + U.

U1 = 0.

K1 =

12

mv21 : U1 = 0.

K2 = 1

2mv2

2 : U2 = mgh.

h = ℓ − ℓcos30".

∴U2 = mg(ℓ − ℓcos30") = mgℓ(1− cos30")

= (2.0 kg)(9.81 m/s2)(3.0 m)(1− cos30")= 7.89 J.

(b) Since energy is conserved,

Re-arranging, we get

(c)

Identify all the forces acting on the mass at :

K2 + U2 = K1 + U1,

i.e., 12

mv22 + mgh = 1

2mv2

1.

v22 = v1

2 − 2gh.

∴v22 = (4.5 m/s)2 − 2(9.81 m/s2)(3.0 m)(1− cos30!)

= 12.36 (m/s)2,i.e., v2 = 3.52 m/s.

T − mgcos30! =

mv22

ℓ.

∴T = mv2

2

ℓ+ gcos30"

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= (2.0 kg)(3.52 m/s)2

3.0m+ (9.81 m/s2)cos30"

⎛

⎝⎜

⎞

⎠⎟

= 25.3 N.

(d) At maximum height: K = 0, so U = K1,

i.e., mghmax = mgℓ(1− cosθmax ) = 12

mv12.

∴cosθmax = 1−v1

2

2gℓ= 1− (4.5 m/s)2

2(9.81 m/s2)(3.0 m)= 0.656,

i.e., θmax = cos−1(0.656) = 49.0".

Question 7.6: A physics student, with mass 80 kg, does a

bungee jump from a platform 100 m above the ground. If

the rubber rope has an un-stretched length of 50 m and

spring constant

(a)  how far above the ground was the student at his lowest

point?

(b) How far above the ground did he achieve his greatest

speed?

(c)  What was his greatest speed?

k = 200 N/m,

Set the zero of gravitational potential energy at the ground.

The total mechanical energy is, therefore, 78,480 J, which

is conserved throughout the jump. At his lowest point

but K + 1

2kℓ2 + mgy" = 78,480 J,

K = 0 and ℓ = 50 − y".

∴100(50 − y!)2 + 80(9.81)y! = 78,480,

i.e., 100y!2 − (10,000 − 784.8)y! + (250,000 − 78,840) = 0.

∴y!2 − 92.15y! +1,715.2 = 0,

i.e., y! =92.15± (92.15)2 − 4(1,715.2)

2.

The solutions are The

appropriate solution is clearly

(b) To find the position where the jumper’s speed is

greatest we need an expression for the kinetic energy (K) in

terms of the vertical position (y) and set to find

where K is a maximum.

If y is the height above the ground where the speed is

greatest and is the amount the rope has stretched then,

since mechanical energy is conserved, at the fastest point

y! = 66.27 m and y! = 25.89 m.

y! = 25.89 m.

dKdy

= 0,

′ℓ

K + 1

2k ′ℓ 2 + mgy = 78,480 J.

Substituting for we get

The extremum occurs when

Check this is a maximum:

(c) when

′ℓ = 50 − y,

K +100(50 − y)2 + 784.8y = 78,480,

i.e., K = 78,480 -100(50 - y)2 − 784.8y

= −171,520 + 9215.2y −100y2

so dKdy

= 9215.2 − 200y.

dKdy

= 0,

i.e., at y = 9215.2

200= 46.08 m.

d2K

dy2= −200, i.e., <0.

Kmax = 78,480 −100(50 − y)2, y = 46.08 m.

∴Kmax = 40,780 J = 1

2mvmax

2

i.e., vmax = 2(40,780 J)

80 kg= 31.9 m/s.

In part (a) we found two possible values for the lowest

point, We chose the

latter, but what does the former correspond to? Note that

both solutions correspond to positions where the kinetic

energy of the jumper is zero. Consider the rope as a

vertical spring and the jumper is the mass at the end of

the spring.

When released the mass oscillates up and down. The

solution is the lowest point of the

oscillation and the solution is the highest

point of the oscillation. The mid-point, at

is where the kinetic energy of the mass is greatest.

y! = 66.27 m and y! = 25.89 m.

y! = 25.89 m

y! = 66.27 m

y = 46.08 m,

When non-conservative forces are involved, such as kinetic

friction, deformation, chemical reactions, etc., we need a

more general statement of the conservation of energy,

which involves the total energy of a system. Take,

for example a bike rider. The

total energy of the system

includes the mechanical energy,

the chemical energy of the rider,

friction, etc., so

Providing there is no input to the system from external

(e.g., wind, a push) then

i.e.,

So an increase in one form of energy (e.g., mechanical) is

compensated by a decrease in another form (e.g., energy

stored by the rider).

Esystem = Ei

i∑ = Emech + Echem + Etherm…

Esystem = constant,

ΔEsystem = 0.

Question 7.7: A block of mass 6.0 kg rests on an

inclined plane. The coefficient of static friction between

the block and the surface of the plane is 0.35. A

gradually increasing force is pulling down on the spring,

which has a spring constant of . If the

angle of the incline is what is the elastic

potential energy of the spring when the block just begins

to move?

θ = 40!,

k = 200 N/m.

x y Fs

FN

mg

fs θ

The potential energy stored in the spring when the block is

about to move is where x is the extension of

the spring. Just before the block moves,

where But what is the force exerted by the

spring? We know that so that the magnitude of

the force associated with that potential function is

U = 1

2kx2,

Fx = Fs∑ − fs − mgsinθ = 0 ... ... [1],and Fy = FN∑ − mgcosθ = 0 ... ... ... [2],

fs = µsFN. Fs,

U = 1

2kx2,

Fs = − dU

dx= kx.

Substituting for in [1] we get

Substituting the given values

Also, the force exerted by the spring, , is

Fs and fs

kx − µsmgcosθ − mgsinθ = 0,

i.e., x =mg(sinθ + µs cosθ)

k.

x = (6.0 kg)(9.81 m/s2)(sin40! + 0.35cos40!)200 N/m

= 0.268 m.

∴U = 12

kx2 = 12

(200 N/m)(0.268 m)2 = 7.18 J.

Fs

Fs = kx = (200 N/m)(0.268)m

= 56.3 N.

Question 7.8: A 2.0 kg block slides down a frictionless

curved ramp, starting from rest at a height of 3.0 m. The

block them slides 9.0 m on a rough horizontal surface

before coming to rest.

(a) What is the speed of the block at the bottom of the

ramp?

(b) How much energy is dissipated by friction?

(c) What is the coefficient of kinetic friction between

the block and the horizontal surface?

UG = 0

(a)  From position to position ,

(b)  The energy dissipated by friction is responsible for

changing the thermal energy of the system. The total

energy is With no external

sources

(K2 + U2) = (K1 + U1).

∴(K2 − K1) = (U1 − U2),

i.e., K2 = mgh = 12

mv22.

∴v2 = 2gh = 2(9.81 m/s2)(3.0 m)

= 7.67 m/s.

Esys = K + UG + Utherm.

ΔEsys = 0.

∴ΔK + ΔUG + ΔEtherm = 0,

i.e., ΔEtherm = −ΔK − ΔUG .

But on the horizontal section

(c) The work done by friction (resulting in ) is

ΔUG = 0 and ΔK = Kf − K2 = 0 − 12

mv22.

∴ΔEtherm = 12

mv22 = 1

2(2.0 kg)(7.67 m/s)2 = 58.9 J.

ΔEtherm

Wf = fkΔx = µkmgΔx = ΔEtherm ,

i.e., µk =ΔEthermmgΔx

= 58.9 J

(2.0 kg)(9.81 m/s2)(9.0 m)= 0.33.

Question 7.9: A block of mass 2.4 kg is dropped onto a

spring with a spring constant from a height

of 5.0 m above the top of the spring.

(a) What is the maximum kinetic energy of the block.

(b) What is the maximum compression of the spring?

This is an interesting problem as I will use different

positions for the zero of gravitational potential energy in

parts (a) and (b).

k = 400 N/m,

(a)  Put at position where the kinetic energy is

a maximum. Then conservation of energy gives

For maximum kinetic energy

UG = 0

K1 + UG1 + UE1 = K2 + UG2 + UE2,

i.e., mg (5.0 m) + Δy( ) = K2 +12

k(Δy)2.

∴K2 = mg (5.0 m) + Δy( )− 12

k(Δy)2.

dK2d(Δy)

= 0 = mg − kΔy,

i.e., Δy = mgk

= (2.4 kg)(9.81 m/s2)400 N/m

= 0.059 m.

But

(b) Now we put at position .

Conservation of energy gives

K2 = mg (5.0 m) + Δy( )− 12

k(Δy)2

= (2.4 kg)(9.81 m/s2)(5.059 m)

− 12

(400 N/m)(0.059 m)2

= 118.4 J.

UG = 0

K1 + UG1 + UE1 = K3 + UG3 + UE3,

i.e., mg (5.0 m) + y!( ) = 12

ky!2.

∴200y!2 − 23.54y! −117.7 = 0,

i.e., y! =23.54 ± (23.45)2 − 4(200)(−117.7)

2(200).

∴y! = 0.828 m or y! = −0.711 m.

Clearly, the former is the physical meaningful solution

and represents the maximum compression of the spring.