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

If a force is applied to an object, the object may experience a change in position, i.e., a displacement.

The work done is W = Fx Δx,

where Fx = Fcosθ and Δx = (x − x! ).

∴W = (ma x )Δx.

But v2 = v!

2 + 2axΔx ⇒ axΔx =12

(v2 − v!2 ).

∴W =

12

mv2 −12

mv!2.

This is known as the work-kinetic energy theorem:

W = Kf − Ki = ΔK.

kinetic energy (K)

When a net force is applied over a distance, say, to move an object, mechanical work is done.

" F

θ

Fx

x x! Δx

Δx

Dimension of work: F ⇒

[M][L]

[T]2 and Δx ⇒ [L].

∴W ⇒

[M][L]2

[T]2(scalar).

Units: N ⋅m⇒ Joule (J).

Dimension of kinetic energy: K ⇒ [M][L]

2

[T]2(scalar).

Note: work and kinetic energy have the same dimensions and units.

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.)

The work-kinetic energy theorem

is very important in physics

When F and Δx are in the same direction, the work is positive and the velocity increases, because the force F

produces a positive acceleration ( a = Fm).

When F and Δx are in opposite directions the work is negative. For example, when kinetic friction is present, the work done by the frictional force is negative, i.e.,

fkΔx = −µkmg( )Δx,

as it tends to slow the motion. So work can be positive or negative.

DISCUSSION PROBLEM [6.1]:

An object initially has kinetic energy K (=

12

mv2 ). If it

now moves in the opposite direction with three times its initial velocity, what is its kinetic energy now?

A: KB: 3KC: −3KD: 9KE: −9K

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 Fy is

= Fy.Δy

= 80.0 N × 3.0 m = 240 J.

(b) The work done by the gravitational force is

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

It is < 0 as the gravitational force is in an opposite direction to the displacement.

(c) The net work done on the box is

Wnet = Fy.Δy − mg.Δy = 240 J −177 J = 63.0 J.

Using the work-KE therorem,

Wnet = ΔK =

12

mv2 −12

mv!2 = 63.0 J,

i.e., v2 =

2 × 63.0 J6.0 kg

= 21.0 m/s( )2,

∴v = 4.58 m/s.

Fy

mg Fy = 80.0 N

6.0 kg

3.0 m

0

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 6.4: Given two vectors

! A = (3ˆ i − 6ˆ j ) and

! B = (−4ˆ i + 2ˆ j ),

calculate

(a) the dot product, ! A •! B , and

(b) the angle between ! A and

! B .

(a) Given ! A = (3ˆ i − 6ˆ j ) and

! B = (−4ˆ i + 2ˆ j ),

! A •! B = AxBx + AyBy.

∴! A •! B = 3× (−4) + (−6) × 2 = −24.

What does a negative product imply?

(b) Also

! A •! B =

! A ! B cosθ⇒ cosθ =

! A •! B

! A ! B

, where θ is

the angle between ! A and

! B . Now

! A = 32 + (−6)2 = 6.71,

! B = (−4)2 + (2)2 = 4.47.

∴θ = cos−1

! A •! B

! A ! B

⎡

⎣ ⎢

⎤

⎦ ⎥ = cos−1 −24

6.71× 4.47⎡ ⎣

⎤ ⎦

= 143".

! A

! B

i

j

θ

Consider something simple like catching a ball.

Note that the ball applies a force on the hand and the hand applies an equal and opposite force on the ball,

i.e., ! F BH = −

! F HB.

Assume that in catching the ball the hand moves through a displacement Δ

! x . Then

WBH =! F BH •Δ! x = FBHΔx,

and WHB =! F HB • Δ! x = −

! F BH • Δ! x = −WBH.

So work can be positive (work done by the hand ON the ball in slowing it down) and negative (work done by the ball ON the hand). Conventionally, when you do work on a system, we take the work as positive work.

Δ! x

! F BH

! F HB

DISCUSSION PROBLEM [6.2]:

In which of the following is zero net work being done?

A: A ball rolling down a hill.B: A physics student stretching a spring.C: A projectile in free fall towards Earth.D: A box being pulled across a rough floor at constant speed.E: A hockey puck slowing down on the ice.

Work done over the small displacement Δxi is FxiΔxi.

∴Wx1→x2 =

Δxi→0Limit FxiΔxii∑( ) = Fx

x1

x2∫ dx,

i.e., area under curve between x1 and x2.

Work done W = Fx Δx = shaded area.

OK for a constant force. But whatif the applied force is not constant?

x2 x1

F

Δxi

Fxi

variable force

Fx

x1 x2

Fx.Δx

Δx

x

F

Alternatively, if the force varies in a simple way with position, we can count squares to find the area or use simple geometry.

For example ...

Fx .

0

a∫ dx

= A1 + A2 + A3.

A1 = 7.5 N ⋅m, A2 = 25 N ⋅m, A3 = 5 N ⋅m.

∴ Total area = 37.5 N ⋅m

Question 6.5: A force, Fx = 0.5x3 N, acts on a object.

What is the work done on the object if it is moved from

x = 1.5 m to x = 3.0 m? Solve the problem (a) using calculus, and (b) graphically.

A1 A2 A3

Fx (N)

x(m) a 0

(a) Plot the function and count squares.

Approximate number of squares:

⇒ 2.5 + 5.75 +10.5 = 18.75But area of a square ⇒ 0.5 m ×1 N = 0.5 N ⋅m

∴W ≈ 18.8 × 0.5 = 9.4 J.

(b) By definition W = Fx

x=1.5

x=3.0∫ dx = 0.5 x3

1.5

3.0∫ dx

=

0.54

x4[ ]1.53.0

= 9.49 J.

Fx (N) Fx = 0.5x3

15

10

5

0 1.0 2.0 3.0 x(m)

Question 6.6: A force, Fx, shown in the plot, acts on a

particle of mass 2 kg, at rest at x = 0.

What is the particle’s speed at x = 10 m?

x(m)

6 4 2 0

0 4 8 2 6 10

10 8

Fx (N)

To determine the final speed we need to determine the work done and then use the work-kinetic energy theorem. The work done is the area under the plot.

The total area is 1[ ]+ [2] + [3]

= 12 +18 +16( ) N ⋅m = 46 N ⋅m.Therefore, using the work-kinetic energy theorem

ΔK =

12

m vf2 − vi

2( ) = 46 N ⋅m.

But vi = 0.

∴vf =

2 × ΔKm

=2 × 46 N ⋅m

2 kg= 6.78 m/s.

x(m)

6 4 2 0

0 4 8 2 6 10

10 8

1

2 3

Fx (N)

Interesting cases: (1) the work done lifting a book ...

The net force on the book is:

! F − m

! g = m

! a y.

The incremental work

done is dW =! F • d! y = Fdy .

So the total work done is

W = Fdy

0

h∫ = (mg + ma y)dy

0

h∫ = mg dy

0

h∫ + m aydy

0

h∫ .

But, m aydy

0

h∫ = m

dvdt0

h∫ dy = m

dydtv"

v∫ dv = m vdv

v"

v∫

=

12

mv2⎡ ⎣ ⎢

⎤ ⎦ ⎥

v"

v= 0, if v" = v (= 0),

i.e., the book starts and finishes at rest (so ΔK = 0).

∴W = mg dy

0

h∫ = mgh .

Therefore, the work done against the gravitational force, does not depend on how the fast/slow the book is lifted, but only on the height!

m! g

h ! F

dy

(2) work done by the gravitational force on a block sliding down a slope ...

As the block slides down the slope, with no friction, the only force on the block in the x-direction is due to the

x −component of the gravitational force:

Fx = mg sinθ.

The work done by the gravitational force on block is:

W = Fx.ℓ = mgℓsin θ.

But ℓsin θ ⇒ h the height of the slope.

∴W = mgh .Note that W is independent of angle θ; the work done by the gravitational force depends only on the vertical height not the length of the actual path!

mg

N

x

y

θθ

ℓ

h

mg

Question 6.7: A force,

! F = 2x2ˆ i + 3y j ( ) N,

where x and y are in meters, acts on a object of mass 5.00 kg.

(a) What is the work done on the object if it is moved from point A, 1 m,2 m( ), to point B, 2 m,3 m( )?

(b) If the speed of the particle at point A is 1.00 m/s, what is its speed when it reaches point B?

(a) The work done from A → B is

WAB =

! F • d! s

A

B∫ =

! F x +

! F y( ) • dxˆ i + dy j ( )

A

B∫

= Fx.dx

xA

xB∫ + Fy .dy

yA

yB∫ .

∴WAB = 2x2.dx

1

2∫ + 3y.dy

2

3∫ = 2

x3

3⎡

⎣ ⎢ ⎤

⎦ ⎥ 1

2

+ 3y2

2⎡

⎣ ⎢ ⎤

⎦ ⎥ 2

3

= 2

8 −13

⎛ ⎝

⎞ ⎠ + 3

9 − 42

⎛ ⎝

⎞ ⎠ = 12.2 J.

(b) Using the work energy theorem:

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

∴KB = 12.2 J +

12

5.00 kg( ) 1.00 m/s( )2 = 14.7 J

=

12

5.00 kg( )vB2.

∴vB =

2 ×14.7 J5.00 kg

= 2.42 m/s.

DISCUSSION PROBLEM [6.3]:

Two wind-driven ice-sleds A and B, with masses M and 2M, respectively, are in a race across a frictionless surface. If the wind exerts a constant force (F) on each boat, which one (if either) wins the race?

A: Sled AB: Sled BC: Ha! It’s a dead-heat!

F

F

M

2MStart Finish

B

A

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: (Re-visiting 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 that the coefficient of kinetic friction was

µk = 0.45 and that 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?

(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

Energy and work are one thing ... but power is quite another.

Look ... one gallon of gas has a certain energy content and so can do a fixed amount of work

~ 120 ×106 J.

But the power produced when the gas burns can have any value ... it depends on how fast it burns!

A gallon of gas will provide enough power to operate a lawnmower for hours (a few HP).

However, a gallon of gas may onlyprovide enough power to operate a jet engine for ~ 1 minute! (~ 2500 HP).

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 motor cycle accelerates from zero to 20 mi/h in 1 s. How long would it take to accelerate from zero to 60 mi/h , assuming the power of the engine is constant?

We have just seen that the power (P) is related to the change in kinetic energy ( ΔK) and time by:

P =

ΔKΔt

.

So, at constant power, Δt ∝ ΔK.

But ΔK =

12

Mvf2 as the motorcycle starts from rest.

• From 0 → 20 mi/h: ΔK1 =

12× M × 202 = 200M.

• From 0 → 60 mi/h: ΔK2 =

12× M × 602 = 1800M.

∴ΔK2 = 9ΔK1,

i.e., Δt2 = 9Δt1.

If Δt1 = 1 s then Δt2 = 9 s.

Question 6.10: A electric motor supplies constant power of 10 hp to the wheels of buggy of mass 500 kg. If all of the power produces motion, how long does it take the buggy to accelerate from

(a) 0 to 10 mi/h,(b) 10 mi/h to 20 mi/h, and(c) 20 mi/h to 30 mi/h?

Conversion factors:

1 hp = 746 W. ∴10 hp = 7.46 kW.

1 mi/h = 0.45 m/s ⇒10 mi/h = 4.5 m/s,

⇒ 20 mi/h = 9.0 m/s,

⇒ 30 mi/h = 13.5 m/s.

We found earlier earlier P =

ΔKΔt

,

i.e., Δt =

ΔKP

=m2P

vf2 − vi

2( ).

=

500 kg2 × 7.46 ×103 W( ) vf

2 − vi2( )

= 3.35 ×10−2 vf2 − vi

2( ).(a) Δt = 3.35×10−2 4.5 m/s( )2 = 0.68 s.

(b) Δt = 3.35×10−2 (9.0 m/s)2 − (4.5 m/s)2( ) = 2.04 s.

(c) Δt = 3.35×10−2 (13.5 m/s)2 − (9.0 m/s)2( ) = 3.39 s.

Question 6.11: A 3.00 kg object starts from rest at

x = 0.50 m and moves along the x -axis under the influence of a single force

Fx = 6.00 + 4.00x − x2,

where Fx is in Newtons and x is in meters.

(a) Find the work done by the force as the object moves from x = 0.50 m to x = 3.00 m.

(b) Find the power delivered by the force as the object passes the point x = 3.00 m.

(a) Since the object is moving along the x-axis, the work done is:

W = 6.00 + 4.00x − x2( )

0.50

3.00∫ .dx

= 6.00x + 4.00 x2

2⎛ ⎝

⎞ ⎠ −

x33

⎡ ⎣

⎤ ⎦ 0.50

3.00

= 6 3.00 − 0.50( ) + 2 3.002 − 0.502( ) − 1

33.003 − 0.503( )⎡

⎣ ⎢ ⎤ ⎦ ⎥

= 23.5 J.

(b) The instantaneous power at any point is

P =! F • ! v .

Since the motion is confined to the x-axis,

P = Fxvx,

where Fx is the force and vx the instantaneous velocity at

x = 3.00 m. To get vx we use the work energy theorem to

find the kinetic energy at x = 3.00 m. Then:

W = ΔK = K at 3 m( ) − K at 0.5 m( ).

Since the object is initially at rest K at 0.5 m( ) = 0,

then W = K at 3 m( ) = 1

2mv x

2

i.e., vx =

2Wm

=2 × 23.5 J3.00 kg

= 3.96 m/s.

∴P = Fxvx

= 6.00 + 4.00 3.0( ) − 3.0( )2( ) N × 3.96 m/s

= 35.6 W.

Fx(at 3 m)

POTENTIAL ENERGY

So far we have seen that when work is done on a isolated object, it leads to a change in kinetic energy. However, there are situations where work does not lead to a change in kinetic energy. We will look at two examples.

1. Springs: consider a spring inside a toy gun.

You compress the spring by applying two equal and opposite forces

! F 1 and

! F 2. The net force on the spring is

zero and so there is no change in the kinetic energy of the spring. However, you have done work on the spring

W =! F 1 • Δ

! x 1 +! F 2 • Δ

! x 2 = 2FΔx,

since ! F 1 =

! F 2 = F and Δ

! x 1 = Δ! x 2 = Δx.

! x 1

! F 1

! F 2

! x 2

So, what happened to the work you did on the spring? Clearly, the shape of the spring has changed as evidenced by its change in length! In fact, when compressed the spring has stored the work you did as potential energy. When released the spring will transfer the stored potential energy to kinetic energy of the ball.

2. Raising an object in the Earth’s gravitational field.

In this scenario we must remember that the object and the Earth are a system; so this is a two-particle system.

However, you are not part of the system; you are an external agent that does work on the

system. The work done by you on the system in lifting the

object from 1 to 2 through a distance h is Wyou = mgh .

But the object starts and finishes at rest, so ΔK = 0.

h mg h 1

2

1

2

Wyou = mgh

Wgrav = −mgh

The object is acted on by two equal and opposite forces - the force you apply upward mg( ) and the force the Earth

applies downward −mg( ). So, the net work done is

Wnet = Wyou + Wgrav = mgh + (−mgh ) = 0,

∴ΔK = 0.But you did work on the system ... if it didn’t increase the kinetic energy of the object what happened to the work?

Note that at 2 the object is capable of doing work; for

example, when released it would fall, increasing its speed, and so it could strike a nail and drive it into the floor! The amount of work the object can do is W = mgh , which is the same as the amount of work you, an external agent, did

raising the object from 1 to 2 .

h mg h 1

2

1

2

Wyou = mgh

Wgrav = −mgh

2

h v

1

When the object is released from rest,

ΔK = W21 = mgh =

12

mv2,

i.e., v = 2gh .

In fact, at 2 the object/Earth system

has stored the work you did as potential energy as it has the potential to return the work, when the object is released.

When the object is raised, the work done by the gravitational force on the object is

W12 = (−mg)h = −mgh .

When it is released, the work done by the gravitational force on the object is:

W21 = +mgh ,

Thus, the net work done by the gravitational force is

2

h v

1

1 2 1 −mgh mgh = 0.

CONSERVATIVE FORCES

A force is called conservative if the total work it does on an object around any closed path is zero ... a closed path is one where the final displacement is zero.

Complementary definition: The work done by a conservative force on an object is zero when the particle returns to its initial position, i.e., when the particle moves around any closed path, no matter the route.

If the work done in going from point 1 to point 2 is W12then (by definition) the work done in going from point 2 to point 1 with a conservative force is W21 = −W12 along any path. The work done by a conservative force along any path is determined only by the end points not the route.

• The gravitational force is a conservative force.• The elastic force is a conservative force.

If you push a book across a table from A to B, you do work against the kinetic frictional force between the book and the table, i.e.,

WAB = Fk.dA→B = mgµkdA→B.Consider two paths a and b. Since b > a, then Wb > Wa . 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 (it produces heat, sound, etc.)

What about non-conservative forces ?

a

b

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

! F = cxˆ j ,

where c is a constant > 0( ). The electron moves in a counterclockwise direction around a square loop in the

xy − plane. If the corners of the loop are at

(0,0), (ℓ,0), (ℓ, ℓ), (0,ℓ),what is the work done on the electron by the force during one complete trip around the loop? Is the force a conservative or non-conservative force?

Along the leg from (0,0) to (ℓ,0) the work done is

W12 =

" F • d" ℓ

0

ℓ∫ .

The force ( " F = cxˆ j ) increases linearly with x, but

" F ⊥d" ℓ ,

along that leg, so the dot product is zero. Similarly for the leg from (ℓ,ℓ) to (0,ℓ). Thus, no work is done on the electron as it travels along the legs parallel to the x-direction, i.e., W12 = W34 = 0.

Along the leg from (ℓ,0) to (ℓ,ℓ) the work done is

W23 =

" F • d" ℓ

0

ℓ∫ = cx j •dy j = cℓ

0

ℓ∫ dy =

0

ℓ∫ cℓ2.

x

y

(0, 0) (ℓ,0)

(ℓ, ℓ) (0, ℓ)

" F

" F

d" ℓ

d" ℓ

d" ℓ d

" ℓ

" F = cℓ j

" F = 0

i j

x

y

(0, 0) (ℓ,0)

(ℓ, ℓ) (0, ℓ)

" F

" F

d" ℓ

d" ℓ

d" ℓ d

" ℓ

" F = cℓ j

" F = 0

Along the leg from (0,ℓ) to (0,0) the force is zero, since

x = 0. So no work is done on the electron along that leg, i.e., W41 = 0. Thus, the total work done on the electron

for the round trip, starting at (0,0) and ending at (0,0), is

W = W12 + W23 + W34 + W41 = cℓ2.

Since the start and end points are the same but the total

work done by the force is non-zero, " F must be a non-

conservative force. Also, since W > 0, the electron hasgained energy as it goes around the loop.

0 0 0

Non-conservative forces and the work-energy theorem

Consider an object falling with air-resistance. There are two forces to consider; the gravitational force (conservative) and the drag force (non-conservative).

The total work done is

W12 = Wgrav + Wnc,

where Wgrav is the work done by the

conservative gravitational force and

Wnc is the work done by the non-

conservative drag force. By the work-energy theorem

W12 = ΔK = K2 − K1.

Also Wgrav = −ΔU = − U2 − U1( ).

∴K2 − K1 = − U2 − U1( ) + Wnc,

i.e., Wnc = K2 + U2( ) − K1 + U1( ),or Wnc = Emech2 − Emech1 = ΔEmech.

Therefore, the work done by a non-conservative force is equal to the change in mechanical energy.

1

2

Conservative forces and the potential energy function

If the potential energy of a system has a unique value at every point

! r , we can define a potential energy function,

U(! r ), which tells us how potential energy varies with

position.

One property of a conservative force is that the work done by the force can be expressed as the difference between the initial and final values of the potential energy, i.e.,

dW =! F • d! s = −dU,

where U is the potential energy function. Hence, the work done by a conservative force equals the decrease in the potential energy from point 1 to point 2. Therefore, the change in potential energy from point 1 to point 2 is

ΔU = U2 − U1 = −

! F •d! s

1

2∫ .

Since from above, dU = −! F • d! s , then, in one-dimension,

dU = −Fx.dx , i.e., Fx = −

dUdx

.

Given that Fx = −

dUdx

, if we know the potential energy

function, U(x), we can determine the force Fx at any

point. Conversely, if we know the functional form of the force Fx, we can find the potential energy function, since

dU = −Fx.dx, i.e., U = − Fx.dx∫ .

Examples ...

[1] Gravitational force:

Let us find the potential energy function for the gravitational force.

We take the y-direction ( j ) vertical. Then, from above, the incremental change in potential energy as the object falls is

dU = −! F • d! s = −(−mgˆ j ) •dy j = mg( )dy.

∴U(y) = mg dy∫ = mgy + U",

where U" is the reference energy when y = 0. Let’s plot

this function:

m! g

j

Note: U⇒ U(y), i.e., it has a unique value at y. This is the potential energy function for the

gravitational force. Normally, we deal only with differences in potential energy, so the choice of U! is entirely arbitrary. Note: the force associated with this

potential energy function is Fz = −

dUdz

= −mg, as

expected.

[2] Elastic force:If you compress a spring a distance x in the x-direction, the force, F1, you exert is given by Hooke’s Law, i.e.,

F1 = kx, where k is the spring constant. The spring does negative work since the force it exerts on you ( F2 = −kx) is opposite to the displacement x.

Height (y)

U(y)

U! ΔU = U2 − U1

2 1

x

F1 = kx F2 = −kx

Therefore, the elastic potential energy function U(x) of the spring is given by:

dU = −! F • d! s = − −F2( ).dx = kx.dx.

∴U(x) = k x.dx =

12∫ kx2 + U".

When there is no displacement, i.e., x = 0, the spring has zero elastic potential energy.

∴U(0) = 0, so U" = 0. At this point, there is no net force acting on the spring (in the x-direction); we say that the spring is in equilibrium. This is the definition of equilibrium, which we first came across in chapter 4 (Newton’s 1st Law).

Equilibrium: At equilibrium

F = 0, i.e., dUdx

= 0,

so U(x) is an extremum. However, we must exercise care

in using that criterion, since dUdx

= 0 does not necessarily

define a minimum in the potential energy function.

x

U(x)

x!

x

U(x)

x!

x

U(x)

x!

Equilibrium position: x!.Stable equilibrium when

U(x! ) is a minimum,

i.e., d2Udx2 > 0.

Equilibrium position: x!.Unstable equilibrium when

U(x! ) is a maximum,

i.e., d2Udx2 < 0.

Equilibrium position: x!.Neutral equilibrium since

U(x! ) is an inflexion point,

i.e., d2Udx2 = 0.

Here are three examples of equilibrium where dUdx

= 0:

Question 6.13: The potential energy function of an object, of mass 2.5 kg, confined to move along the x-axis, is

U(x) = 3x2 − 2x3,

where U(x) is in Joules and x is in meters. If the only force acting on the object is the force associated with this potential energy function,

(a) plot the force as a function of position.

(b) At what positions is the object in equilibrium and what type of equilibrium is associated with each position?

(c) Confirm the result by plotting the potential energy function.

(a) The force associated with the potential function is

F(x) = −

dU(x)dx

= −6x + 6x2.

(b) At equilibrium F(x) = 0,

i.e., 6x2 − 6x = 6x(x −1) = 0

Therefore, the equilibrium positions are at x = 0 and

x = 1 m. To find the type of equilibrium look at the signs

of the seond derivative,

d2U(x)dx2 , at x = 0 and x = 1 m.

Since U(x) = 3x2 − 2x3, then

d2U(x)dx2 = 6 −12x.

• At x = 0:

d2U(x)dx2 = 6 (> 0) ⇒ Stable.

• At x = 1 m:

d2U(x)dx2 = −6 (< 0) ⇒ Unstable.

F(x)

x(m) −1 1 2

(c)

The plot of the potential function has • a minimum at x = −1 m where the object is in stable equilibrium, and• a maximum at x = 1 m where the object is in unstable equilibrium.

x(m)

U(x) = 3x2 − 2x3 5

−5

−1 1 2

U(J)