CHAPTER 6

WORK AND ENERGY

• Work and kinetic energy ! work-kinetic energy theorem

• Work done by a variable force ! the dot product

• Power

• Potential energy ! conservative forces ! non-conservative forces ! conservative forces and the potential-energy function ! equilibrium

Work and kinetic energy

If a net force is applied on an object, the object may

experience a change in position, i.e., a displacement.

When a net force is applied over a distance, work is done.

The work done is where the force is in the

direction of the displacement, In the lower

figure,

But

kinetic energy (K)

This is known as the work-kinetic energy theorem

W = FxΔx, Fx

Δx = x − x!.

Fx = Fcosθ.

∴W = (max )Δx,

v2 = v!

2 + 2axΔx ⇒ axΔx = 12

(v2 − v!2).

∴W = 1

2mv2 − 1

2mv!

2.

W = Kf − Ki = ΔK.

Dimension of work:

(scalar).

Units:

NOTE:

• Kinetic energy and work have the same dimension

and units.

• Work can be positive or negative. If a force

increases the speed of an object, the work done by

the force is positive. If a force decreases the speed

of an object (e.g., frictional force) the work done by

the force is negative.

N ⋅m and/or Joule (J).

F ⇒ [M][L]

[T]2 and Δx ⇒ [L].

∴W ⇒ [M][L]2

[T]2 Question 6.1: A constant force of 10 N acts on a box of mass 2.0 kg for 3.0 s. If the box was initially at rest and the coefficient of kinetic friction between the box and the surface is µk = 0.3,

(a) what is its speed of the box after 3.0 s?

(b) How far did the box travel?

(c) What was the work done by the applied force?

(d) How much work was done by the frictional force?

(a) The net force acting on the box is

Fnet = F − fk = F − µkN = F − µkmg

= 10 N − (0.3× 2.0 kg × 9.81 m/s2) = 4.11 N.

Therefore, the acceleration of the box is

a = Fnet

m = (4.11 N)(2.0 kg) = 2.06 m/s2.

The final velocity after 3.0 s is given by v = v! + at

= 2.06 m/s2 × 3.0s = 6.18 m/s.

(b) To find the distance traveled we have:

v2 = v!2 + 2aΔx,

i.e., Δx =

v2 − v!2

2a=

6.18 m/s( )2

2 × 2.06 m/s2 = 9.27 m.

Δx

F = 10N

v! = 0 v = ?

m = 2.0 kg

t = 3.0 s mg

fk F

N (c) The work done by the applied force is W = F.Δx

= (10 N)(9.27 m) = 92.7 N ⋅m.

(d) The work done by the frictional force is

Wf = −µkmg.Δx

= −(0.3)(2.0 kg)(9.81 m/s2)(9.27 m)

= −54.6 J (or N ⋅m).

Note the net work done by the applied force and the frictional force is

Wnet = (92.7 − 54.6) = 38.1 J.

It is the net work that produces the increase in the speed of the box as we can see from the work-kinetic energy theorem.

Wnet = ΔK =

12

mv2 −12

mv!2

=

12(2.0 kg)(6.18 m/s)2 = 38.2 J. (Rounding error.)

Question 6.2: Which of the following requires the most work to be done by the engine of a motor cycle?

A: Accelerating from 60 km/h to 80 km/h.B: Accelerating from 40 km/h to 60 km/h.C: Accelerating from 20 km/h to 40 km/h.D: Accelerating from 0 to 20 km/h.E: The same in each case.

Let the mass of the motorcycle and rider be M. The work done is given by the work-kinetic energy theorem, i.e.,

W = ΔK =

12

M vf2 − vi

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ .

For A: ΔK =

12

M 802 − 602( ) = 1400M.

For B: ΔK =

12

M 602 − 402( ) = 1000M.

For C: ΔK =

12

M 402 − 202( ) = 600M.

For D: ΔK =

12

M 202( ) = 200M.

So case A requires most work by the engine. Even though the change in speed is the same in each case, the work done is greatest for A because the initial speed is the greatest also.

Question 6.3: A box, of mass 6.0 kg, is raised a distance of 3.0 m from rest by a vertical force of 80.0 N. Find

(a) the work done by the force, (b) the work done by the gravitational force, and (c) the final velocity of the box.

(a) The work done by the applied

force is

(b) The work done by the gravitational force is

The work is negative because the gravitational force is in

the opposite direction to the displacement.

(c) The net work done on the box is

Using the work-kinetic energy theorem,

But

Fy = 80 N 3 m

Fy

mg 6.0 kg

Fy

FyΔy = (80 N)(3.0 m) = 240 J.

(−mg)Δy = −(6.0 kg)(9.81 m/s2)(3.0 m) = −177 J.

Wnet = (Fy − mg)Δy = 240 J −177 J = 63 J.

Wnet = ΔK = 1

2mv2 − 1

2mv!

2 = 63 J,

v! = 0.

∴v2 = 2(63 J)6.0 kg

= 21.0 (m/s)2,

i.e., v = 4.58 m/s.

Question 6.4: A car accelerates from rest and gains a

certain amount of kinetic energy. The Earth …

A: gains more kinetic energy.

B: gains the same amount of kinetic energy.

C: gains less kinetic energy.

D: loses kinetic energy as the car gains it.

The work-kinetic energy theorem tells us that the kinetic

energy gained is equal to the work done, i.e., the product

of the force and the displacement. By Newton’s 3rd

Law, the force exerted by the car on the Earth is equal

and opposite to the force exerted by the Earth on the car.

However, the distance over which these forces act are

not equal; the Earth moves a negligible amount as a

result of the cars motion. Therefore, the answer is

C: the Earth gains less kinetic energy.

If F is not parallel to the displacement, we have to use the component of the force parallel to the displacement:

i.e., the work done W = (! F cosθ)Δ! s =

! F Δ! s cosθ.

The dot or scalar product of

two vectors, ! A and

! B is

defined as

! A •! B =

! A ! B cosθ.

If ! A ⇒ (Ax,Ay, Az ) and

! B ⇒ (Bx,By,Bz )

Then ! A •! B = AxBx + AyBy + AzBz

~ see revision notes on website ~

Therefore, the work done is W = ! F • Δ! s

θ ! A

! B

! F

θ

Δ! s Question 6.5: A force acts on an

object and produces a displacement

Find

(a) the work done by the force, and

(b) the angle between

!F = (2i − 2 j+ 2k) N

Δ!s = (2i + j− 3k) m.

!F and Δ!s.

(a) The work done by the force is

What is the significance of the negative sign?

(b) We have

where is the angle between Now

Since

W =!FiΔ!s = FxΔsx + FyΔsy + FzΔsz

= (2 N)(2 m) + (−2 N)(1 m) + (2 N)(−3 m)= −4 J.

!FiΔ!s =

!F Δ!s cosθ, i.e., cosθ =

!FiΔ!s!F Δ!s

,

θ !F and Δ!s.

!F = (2 N)2 + (−2 N)2 + (2 N)2 = 3.46 N.

Δ!s = (2 m)2 + (1 m)2 + (−3 m)2 = 3.74 m.

∴θ = cos−1

!FiΔ!s!F Δ!s

⎛

⎝⎜

⎞

⎠⎟ = cos−1 −4 J

(3.46 N)(3.74 m)⎛⎝⎜

⎞⎠⎟

= 108!.

90! < θ < 270!, then W < 0.

In the preceding, the applied force has been constant. How

do we handle the situation if the force varies?

Work done over the small displacement is

i.e., the area under the curve between

Δxi Fxi

Δxi.

∴Wx1→x2

= Δxi→0Limit Fxi

Δxii∑( ) = Fxx1

x2∫ dx,

x1 and x2.

W = FxΔx ⇒ shaded area.

Question 6.6: The force shown above acts on an

object of mass 2 kg. If the object is at rest at what

is its speed at

Fx

x = 0,

x = 10 m?

We find the work done over the displacement and then

use the work-kinetic energy theorem to determine the

speed.

The work done is the area under the plot,

i.e.,

Therefore, using the work-kinetic energy theorem

But

[1]+ [2]+ [3] = (12 +18+16) J = 46 J.

ΔK = 1

2m(v2 − v!

2) = 46 J.

v! = 0.

∴v = 2ΔK

m= 2(46 J)

2 kg= 6.78 m/s.

Question 6.7: A force,

where x, y and z are in meters, acts on an object of mass

5.00 kg.

(a) How much work is done on the object if it moves

from point A, (1 m,2 m,1m), to point B, (2 m,3 m,2m)?

(b) If the speed of the object at point A is 1.00 m/s,

what is it speed at point B?

!F = 2x2i + 3yj− 2zk( ) N,

(a)  The work done from A to B is

(b) Using the work-kinetic energy theorem:

WAB =

!Fid!s = Fxi + Fy j+ Fzk( )

A

B∫

A

B∫ i dxi + dyj+ dzk( )

= Fx

xA

xB∫ dx + Fy

yA

yB∫ dy + Fz

zA

zB∫ dz.

WAB = ΔK = KB − KA, i.e., KB = KA + WAB.

∴KB = 1

2(5.00 kg)(1.00 m/s)2 + 9.2 J = 11.7 J

= 1

2(5.00 kg)vB

2.

∴vB = 2(11.7 J)

5.00 kg= 2.16 m/s.

∴WAB = 2x2

1

2∫ dx + 3y

2

3∫ dy − 2z

1

2∫ dz

= 2 x3

3⎡⎣⎢

⎤⎦⎥1

2+ 3 y2

2⎡

⎣⎢

⎤

⎦⎥

2

3

− 2 z2

2⎡⎣⎢

⎤⎦⎥1

2

= 28−1

3⎛⎝⎜

⎞⎠⎟+ 3

9 − 42

⎛⎝⎜

⎞⎠⎟− 2

4 −12

⎛⎝⎜

⎞⎠⎟= 9.2 J.

POWER

Power is the rate at which work is done or energy is dissipated (i.e., quickly or slowly). For example,

running up a flight of stairs quickly requires more power than walking up the same flight of stairs slowly. At any instant, the instantaneous power is

P =

dWdt

=! F •d! s

dt=! F •

d! s dt

=! F • ! v (scalar),

so power can be > 0 or < 0 depending on the angle between

! F and

! v .

Dimensions: Power ⇒

work donetime

⇒[M][L]2

[T]2[T]

=[M][L]2

[T]3.

Units: J/s ⇒ watts (W)1000 W ⇒ 1 kW1 HP ⇒ 746 W

Question 6.8: (Revisiting question 5.5.) The driver of a

1200 kg car moving at 15.0 m/s is forced to slam on the

brakes. The car skids to a halt after traveling a distance of

25.5 m.

We found the coefficient of kinetic friction was

and it took 3.40 s for the car to stop.

(a) How much work is done by the frictional force?

(b) What is the average power dissipated by the

frictional force?

µk = 0.45

(a) Two ways to find the work done by the frictional force:

(i) Wf = −fk.ℓ = −(µkmg)ℓ

= −0.45(1200 kg)(9.81 m/s2)(25.5 m)

= −1.35 ×105J.

(ii) Wf = ΔK =

12

mv2 −12

mv"2

= −

12

(1200 kg)(15.0 m/s)2 = −1.35×105J.

(b) Average power dissipated by the frictional force is

Pav =

Wft

=−1.35×105J

3.40 s

= −3.97 ×104 W = −39.7 kW.

The power is < 0 as Wf < 0, which means that energy is

removed from the system (as the car is slowing down).

fk

mg

N

x" x

v"

ℓ

v = 0

One-dimensional example:

P =! F • ! v = Fxvx = ma xvx

i.e., ax =

Pmv x

.

So, if the engine operates with constant power output, the resulting acceleration is inversely proportional to the velocity, i.e., as v increases, a decreases. Also

P = Fxvx = maxvx = mv x

dvxdt

=

ddt

12

mvx2⎛

⎝ ⎜ ⎞ ⎠ ⎟ =

dKdt

,

i.e., the instantaneous power at some time t is the rate of change of kinetic energy at that same time.

Question 6.9: A car, with mass 1000 kg, accelerates from

zero to 20 mi/h in 2 s. Assuming the power of the engine

is constant

(a) how long would it take to accelerate from zero to

60 mi/h?

(b) What is the power generated by the engine?

(a) Power is related to the change in kinetic energy over

a time interval , viz:

So, at constant power,

Also, since where M is the mass of the

car, and v is the final speed.

• From 0 to 20 mi/h:

• From 0 to 60 mi/h:

Since

So, the time required to increase the speed by a factor of n,

increases by

(b) Convert 20 mi/h to m/s:

ΔK

Δt

P = ΔK

Δt.

Δt ∝ΔK.

v! = 0, ΔK = 1

2Mv2,

ΔK1 =

12

M(20)2 = 200M.

ΔK2 = 1

2M(60)2 = 1800M.

∴ΔK2 = 9ΔK1, i.e., Δt2 = 9Δt1.

Δt1 = 2 s then Δt2 = 18 s.

n2.

⇒ (20 mi/h)(1.61×103 m/mi)

3600 s/h= 8.94 m/s.

Therefore, the power generated by the engine is

Check:

P =ΔK1Δt1

= 12

(1000 kg)(8.94 m/s)2

2 s

= 2.00 ×104 W ⇒ 26.8 HP.

60 mi/h ⇒ (60 mi/h)(1.61×103 m/mi)3600 s/h

= 26.8 m/s.

∴ΔK2 = 12

(1000 kg)(26.8 m/s)2 = 3.59 ×105 J.

∴P =ΔK2Δt2

= 3.59 ×105 J18 s

= 2.0 ×104 W ⇒ 26.8 HP.

Question 6.10: A force,

acts on a 3.00 kg object. If the object is initially at rest,

what is the instantaneous power delivered by the force after

2 s?

!F = (5.40i + 2.70 j) N,

The acceleration of the object is

The velocity after 2 s is

so the instantaneous power is

!a =!Fm

= (5.40i + 2.70 j) N3.00 kg

= (1.80i + 0.90 j) m/s2.

!v = !v" +

!at = (1.80i + 0.90 j)(2 s) = (3.60i +1.80 j) m/s,

P =!Fi!v = (5.40i + 2.70 j)i(3.60i +1.80 j) W= (5.40 × 3.60) + (2.70 ×1.80) W

= 24.3 W.

Question 6.11: A 5.00 kg object starts from rest at

and moves along the x-axis under the influence of a single

force

where is in Newton’s and x is in meters.

(a) How much work is done as the object moves from

to

(b) What power is delivered by the force as the object

passes the point

x = 0

Fx = x3 − 2.00x2 + 3.00,

Fx

x = 0 x = 3 m?

x = 3 m?

(a)  The work done is

(b) The instantaneous power is We find

using the work-kinetic energy theorem:

W = (x3 − 2.00x2 + 3.00)dx0

3∫

= x4

4− 2.00

x3

3

⎛

⎝⎜

⎞

⎠⎟ + 3.00x

⎡

⎣⎢⎢

⎤

⎦⎥⎥0

3

= 11.3 J.

P =!Fi!v ⇒ Fxvx.

vx

W = ΔK = Kf − Ki =12

mvx2.

∴vx = 2Wm

= 2(11.3 J)5.00 kg

= 2.13 m/s.

Fx = (3.00 m)3 − 2.00(3.00 m)2 + 3.00

= 12.0 N.∴P = (12.0 N)(2.13 m/s) = 25.6 W.

Potential energy

Consider lifting an object through

a distance h; how much work is

done? The net force is

The work done by you in lifting

the object through a distance dy is so the

total work done is

But

i.e., the object starts and finishes at rest.

So the work done by you against the gravitational force

depends only on h and not on the rate the work was done.

!F − m!g = m!ay.

dW =!Fid!y = Fdy,

W = F

0

h∫ dy = (mg + m!ay)dy = mg dy

0

h∫

0

h∫ + m ay dy

0

h∫ .

m ay dy0

h∫ = m dv

dt0

h∫ dy = m dy

dtv!

v∫ dv = m vdv

v!

v∫

= 12

mv2⎡

⎣⎢

⎤

⎦⎥

v!

v= 0, if v! = v ( = 0),

∴W = mg dy

0

h∫ = mgh.

What about lifting the object at an angle to the horizontal?

The net force along the

sloped line (s) is

where is the acceleration

along s and is the component of the weight force.

But, if the object starts and finishes at rest, the second

integral is zero, as we showed on the previous slide. Also,

since and so the work you do is

i.e., the same as before, so the work done is independent of

the route!

θ

θ mg

F

h

ℓ

F − mgsinθ = mas,

as

∴W = (mgsinθ + mas)dℓ = mg sinθdℓ

0

ℓ∫

0

ℓ∫ + m as

0

ℓ∫ dℓ.

h = ℓsinθ then dh = sinθdℓ,

W = mg dh

0

h∫ = mgh,

mgsinθ

Although we have done work on the object in both cases

there is no change in kinetic energy. So, what’s happened

to the work? Clearly the configuration has changed since

the object is in a different position after the work was

done. In fact, the gravitational potential energy of the

object has changed. The amount of potential energy

gained is equal to the amount of work we do, i.e.,

When the object is raised, the work done by

the gravitational force is

In the raised position, the object is capable of doing work;

for example, when released and allowed to fall, its kinetic

energy increases so it could strike a nail and drive it into

the floor! The work done by the gravitational force when

the object falls to its original position, from [2] to [1], is

(ΔUG )

ΔUG = (U2 − U1) = mgh.

W12 = (−mg)h = −ΔUG = −(U2 − U1).

W21 = (−mg)(−h) = mgh = (U2 − U1).

F

mg [1]

[2]

So the net work done by the gravitational force is zero, i.e.,

The gain in kinetic energy of the object as it falls is

i.e.,

W21 = −W12.

ΔK = W21 = mgh

ΔK = W21 = mgh,

12

mv2 = mgh ⇒ v = 2gh.

A similar situation arises when you compress a spring,

e.g., in a toy gun. The spring is compressed by applying

two equal and opposite forces. The net force on the spring

is zero so there’s no change in kinetic energy, but you have

done work on the spring. What’s happened to the work?

Clearly, the length of the spring has changed. In fact, the

spring has stored the work you did as elastic potential

energy. The increase in potential energy is

where k is the spring constant – a measure of the stiffness

– and x is the change in length of the spring. When

released the spring will transfer the potential energy

gained to the kinetic energy of the ball. The net work

done by the elastic force in this example is, therefore, zero.

ΔUE = 1

2kx2,

Conservative forces

If the work done by a force in going from configuration 1

to configuration 2 is and the work done in going

from configuration 2 to configuration1 is then, if

the force is defined as a conservative force. The

gravitational and the elastic forces are examples of

conservative forces.

Complementary definition: The work done on an object

by a conservative force is zero when the object returns to

its initial position, i.e., when the object moves around any

closed path, no matter the route.

W12

W21,

W21 = −W12

What about non-conservative forces?

Consider pushing a book across a table from position A

to position B. You have to do work against the kinetic

frictional force between the book and table. Then

Consider two different paths, a and b. Since then

So, the work done against the frictional force

does depend on the path. Therefore, the frictional force

is a non-conservative force.

NOTE: the work done by a non-conservative force is

non-recoverable (producing heat, sound, etc.).

WAB = FkdA→B = mgµkdA→B.

b > a,

Wb > Wa.

Question 6.12: In a certain region of space, the force on

an electron is given by

where c is a constant. The electron moves round a

square loop in the x-y plane. If the corners of the loop

are at

how much work done on the electron by the force while

completing one trip around the loop? Is the force a

conservative force?

!F = cyi,

(0,0), (ℓ,0), (ℓ,ℓ), (0,ℓ),

(0,0) (ℓ,0)

(ℓ,ℓ) (0,ℓ)

x

y

dℓj dℓj

dℓi

dℓi

!F = 0

!F = cℓi

!F

!F

• From the force

• From the force increases linearly but

so Similarly,

• From

So, the total work done around the loop is

Since the start and end points are the same, the work done

by the force around the loop is non-zero. Therefore, is

a non-conservative force. Note, if the electron gains

energy as it goes around the loop.

(0,0) to (ℓ,0) F = 0. ∴W12 = 0.

(ℓ,0) to (ℓ,ℓ)

!F ⊥ d

!ℓ, W23 = 0. W41 = 0.

(ℓ,ℓ) to (0,ℓ)

W34 =

!Fid!ℓ = cℓiidxi = cℓ dx = −c

ℓ

0∫

ℓ

0∫

ℓ

0∫ ℓ2.

W = W12 + W23 + W34 + W41 = −cℓ2.

!F

c < 0

Potential energy function

A property of a conservative force is that there is a

potential energy function associated with the force,

which tells us how potential energy varies with position.

As we’ve seen with the gravitational force, the work done

can be expressed as the difference between the initial and

final values of the potential energy, i.e.,

Writing this equation in incremental form

where U is the potential energy function. Therefore, the

change in potential energy from point A to point B is

Since then in one-dimension,

U(r)

WA→B = −ΔU = −(UB − UA ).

dW = −dU =!Fid!s,

ΔU = UB − UA = dU

A

B∫ = −

!Fid!s.

A

B∫

dU = −!Fid!s,

dU = −Fxdx i.e., Fx = − dU

dx.

What this means is, if we know the functional form of

U(x), we can determine the force at any point.

Conversely, if we know the functional form of the force

we can find the potential energy function since

As an example, the potential energy function for a

compressed spring is

Therefore, the force associated with this potential energy

function is

which is Hooke’s Law, where x is the compressed distance.

When the spring is relaxed and This

is known as equilibrium (chapter 4).

At equilibrium so U(x) is an extremum.

Fx

Fx ,

dU = −Fxdx i.e., U = − Fx dx∫ .

U(x) = 1

2kx2.

F(x) = − dU(x)

dx= −kx,

x = 0, F = 0, dUdx

= 0.

dUdx

= 0,

However, does not uniquely define a minimum in

the potential energy function.

STABLE EQUILIBRIUM Equilibrium position

is a minimum.

UNSTABLE EQUILIBRIUM Equilibrium position

is a maximum.

NEUTRAL EQUILIBRIUM Equilibrium position

is a an inflexion point.

dUdx

= 0

x!.

U(x!) d2U

dx2> 0

d2U

dx2< 0

d2U

dx2= 0

x!.

U(x!)

x!.

U(x!)

Question 6.13: A potential energy function varies as

where U is in Joules and x in meters.

(a) At which point(s) is this object in equilibrium?

(b) What is the state of equilibrium at these positions?

U(x) = 2x4 − x2,

(a)  Equilibrium occurs where

therefore, the equilibrium points are solutions of the

equation:

The solutions are

(b) For the state of equilibrium, find the sign of

When

When

When

F = − dU

dx= 0.

− dU

dx= − 8x3 − 2x( ),

8x3 − 2x( ) = 2x 4x2 −1( ) = 0.

x = 0 and x = ±0.5 m.

x = −0.5 m,

d2U

dx2= 24x2 − 2.

d2U

dx2= 6 − 2 = 4, i.e., > 0 (stable).

x = 0,

d2U

dx2= 6 − 2 = 4, i.e., > 0 (stable).

d2U

dx2= −2, i.e., < 0 (unstable).

x = +0.5 m,

Stable Stable Unstable

U(x) = 2x4 − 2x