If a force is applied to an object, the object may
experience a change in position, i.e., a displacement.
CHAPTER 6
When a net force is applied over a distance, say,
to move an object, mechanical work is done.
Fx
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 potentialenergy function
! equilibrium
Δx
x!
"
F
x
θ
Δx
The work done is W = Fx Δx,
where Fx = F cosθ and Δx = (x − x ! ).
∴W = (ma x )Δx.
1
2
2
But v 2 = v ! + 2a x Δx ⇒ a x Δx = (v 2 − v ! ).
2
1
1
2
∴W = mv 2 − mv ! .
2
2
kinetic energy (K)
This is known as the work-kinetic energy theorem:
W = K f − Ki = ΔK.
Dimension of work:
F⇒
∴W ⇒
[M][L]
[T]2
[M][L]2
[T]2
and Δx ⇒ [L].
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
(scalar).
the surface is µ k = 0.3,
Units: N ⋅ m ⇒ Joule (J).
(a) what is its speed of the box after 3.0 s?
[L]2
Dimension of kinetic energy: K ⇒ [M] 2 (scalar).
[T]
(b) How far did the box travel?
Note: work and kinetic energy have the same dimensions
and units.
(c) What was the work done by the applied force?
(d) How much work was done by the frictional
force?
F = 10 N
m = 2.0 kg
v! = 0
Δx
v=?
(c) The work done by the applied force is W = F.Δx
N
fk
t = 3.0 s
= (10 N)(9.27 m) = 92.7 N ⋅ m.
F
mg
(d) The work done by the frictional force is
Wf = −µ k mg.Δx
(a) The net force acting on the box is
Fnet = F − f k = F − µ k N = F − µ k mg
= −(0.3)(2.0 kg)(9.81 m/s 2 )(9.27 m)
= 10 N − (0.3 × 2.0 kg × 9.81 m/s 2 ) = 4.11 N.
= −54.6 J (or N ⋅ m).
Therefore, the acceleration of the box is
F
a = net m = (4.11 N) (2.0 kg) = 2.06 m/s2 .
Note the net work done by the applied force and the
frictional force is
The final velocity after 3.0 s is given by v = v ! + at
= 2.06 m/s2 × 3.0s = 6.18 m/s.
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
(b) To find the distance traveled we have:
v 2 = v !2 + 2aΔx,
i.e., Δx =
v 2 − v !2
(6.18 m/s)2
=
= 9.27 m.
2a
2 × 2.06 m/s2
theorem.
1
1
Wnet = ΔK = mv 2 − mv !2
2
2
1
= (2.0 kg)(6.18 m/s) 2 = 38.2 J. (Rounding error.)
2
The work-kinetic energy theorem
is very important in physics
DISCUSSION PROBLEM [6.1]:
1
An object initially has kinetic energy K (= mv 2 ). If it
2
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 = F m).
now moves in the opposite direction with three times its
initial velocity, what is its kinetic energy now?
A: K
B: 3K
When F and Δx are in opposite directions the work is
C: −3K
negative. For example, when kinetic friction is present,
D: 9K
the work done by the frictional force is negative, i.e.,
E: −9K
f kΔx = ( −µ k mg )Δx,
as it tends to slow the motion. So work can be positive or
negative.
Let the mass of the motorcycle
and rider be M. The work done is
given by the work-kinetic energy
theorem, i.e.,
1 ⎛ 2
2⎞
W = ΔK = M⎜ v f − v i ⎟ .
⎠
2 ⎝
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.
(
)
(
)
(
)
1
For A: ΔK = M 802 − 602 = 1400M.
2
1
For B: ΔK = M 602 − 402 = 1000M.
2
1
For C: ΔK = M 402 − 202 = 600M.
2
1
For D: ΔK = M 202 = 200M.
2
( )
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.
Fy
Fy = 80.0 N
6.0 kg
3.0 m
mg
(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
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.
(−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,
0
1
2 1
2
Wnet = ΔK = mv − mv ! = 63.0 J,
2
2
2 × 63.0 J
i.e., v 2 =
= 21.0 ( m/s)2 ,
6.0 kg
∴v = 4.58 m/s.
If F is not parallel to the displacement, we have to use
the component of the force parallel to the displacement:
!
F
θ
!
Δs
!
! ! !
i.e., the work done W = ( F cosθ) Δ s = F Δ s cosθ.
!
B
θ
!
A
The dot or scalar product of
!
!
two vectors, A and B is
defined as
! !
! !
A • B = A B cosθ.
!
!
If A ⇒ (A x , A y , A z ) and B ⇒ (B x ,B y ,B z )
! !
Then A • B = A xB x + A yB y + A zB z
~ see revision notes on website ~
!
!
Therefore, the work done is W = F • Δ s
Question 6.4: Given two vectors
!
!
A = (3ˆi − 6 ˆj) and B = (−4 ˆi + 2 ˆj),
calculate
! !
(a) the dot product, A • B , and
!
!
(b) the angle between A and B.
!
!
(a) Given A = (3ˆi − 6 ˆj) and B = (−4 ˆi + 2 ˆj),
! !
A • B = A xB x + A yB y .
! !
∴ A • B = 3 × (−4) + (−6) × 2 = −24.
Consider something simple like catching a ball.
!
FHB
!
FBH
What does a negative product imply?
! !
! !
! !
A•B
(b) Also A • B = A B cosθ ⇒ cosθ = ! ! , where θ is
AB
!
!
the angle between A and B. Now
!
A = 32 + (−6)2 = 6.71,
!
B = (−4)2 + (2)2 = 4.47.
! !
⎡
A
• B⎤
−24 ⎤
∴θ = cos −1 ⎢ ! ! ⎥ = cos −1 ⎡
= 143" .
A
B
⎣
6.71
×
4.47
⎦
⎣
⎦
!
Δx
Note that the ball applies a force on the hand and the
hand applies an equal and opposite force on the ball,
!
!
i.e., FBH = − FHB .
Assume that in catching the ball the hand moves through
!
a displacement Δx . Then
!
!
WBH = FBH • Δx = FBH Δx,
!
!
!
!
and WHB = FHB • Δx = −FBH • Δx = −WBH .
ˆj
!
B
ˆi
θ
!
A
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.
OK for a constant force. But what
if the applied force is not constant?
F
Δx
Fx
Fx .Δx
DISCUSSION PROBLEM [6.2]:
Work done W = Fx Δx = shaded area.
In which of the following is zero net work being done?
A: A ball rolling down a hill.
F
variable force
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
x
x2
x1
Fx i
constant speed.
E: A hockey puck slowing down on the ice.
Δx i
x1
x2
Work done over the small displacement Δx i is Fx i Δx i .
(
)
x2
∴Wx1 → x 2 = Limit ∑ i Fx i Δx i = ∫ Fxdx,
Δx i →0
i.e., area under curve between x1 and x 2.
x1
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 ...
(a) Plot the function and count squares.
15
Fx (N)
Fx = 0.5x 3
10
Fx (N)
a
5
∫ Fx .dx
0
= A1 + A2 + A3.
0
A1
A2
A3
a
x(m)
A1 = 7.5 N ⋅ m, A2 = 25 N ⋅ m, A3 = 5 N ⋅ m.
∴ Total area = 37.5 N ⋅ m
1.0
0
2.0
3.0
x(m)
Approximate number of squares:
⇒ 2.5 + 5.75 + 10.5 = 18.75
But area of a square ⇒ 0.5 m × 1 N = 0.5 N ⋅ m
∴W ≈ 18.8 × 0.5 = 9.4 J.
Question 6.5: A force, Fx = 0.5x 3 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.
(b) By definition W =
=
x=3.0
∫ Fx
x=1.5
[ ]
3.0
dx = 0.5 ∫ x 3dx
1.5
0.5 4 3.0
x
= 9.49 J.
1.5
4
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.
Fx (N)
Question 6.6: A force, Fx, shown in the plot, acts on a
10
8
6
4
2
0
particle of mass 2 kg, at rest at x = 0.
Fx (N)
10
8
6
4
2
0
x(m)
0 2 4 6 8 10
What is the particle’s speed at x = 10 m?
2
1
3
x(m)
0 2 4 6 8 10
The total area is [1] + [2] + [3]
= (12 + 18 + 16 ) N ⋅ m = 46 N ⋅ m.
Therefore, using the work-kinetic energy theorem
1
ΔK = m v f 2 − v i2 = 46 N ⋅ m.
2
(
)
But v i = 0.
∴v f =
2 × ΔK
2 × 46 N ⋅ m
=
= 6.78 m/s.
m
2 kg
Interesting cases: (1) the work done lifting a book ...
!
F
h
The net force on the book is:
!
!
!
F − mg = ma y .
dy
The incremental work
! !
done is dW = F • dy = Fdy .
!
mg
So the total work done is
h
h
h
h
0
0
0
0
W = ∫ Fdy = ∫ (mg + ma y )dy = mg ∫ dy + m ∫ a ydy .
h
(2) work done by the gravitational force on a block
sliding down a slope ...
h dv
v
v
dy
But, m∫ a ydy = m ∫ dy = m ∫ dv = m ∫ vdv
0
0 dt
v dt
v
"
"
⎤v
⎡1
= ⎢ mv 2 ⎥ = 0, if v " = v (= 0),
⎣2
⎦ v"
i.e., the book starts and finishes at rest (so ΔK = 0).
h
∴W = mg ∫ dy = mgh .
0
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!
N
ℓ
y
x
h
θ
θ
mg
mg
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 sin θ.
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!
(a) The work done from A → B is
B!
!
! B !
WAB = ∫ F • d s = ∫ Fx + Fy • dxˆi + dyˆj
Question 6.7: A force,
!
F = 2x 2 ˆi + 3yˆj 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
A
xB
yB
xA
yA
)
= ∫ Fx .dx + ∫ Fy .dy .
2
3
⎡ x3 ⎤
⎡ y2 ⎤
∴WAB = ∫ 2x .dx + ∫ 3y.dy = 2 ⎢ ⎥ + 3⎢ ⎥
1
2
⎣ 3 ⎦1
⎣ 2 ⎦2
2
2
3
8 − 1⎞ ⎛ 9 − 4 ⎞
= 2⎛
+3
= 12.2 J.
⎝ 3 ⎠ ⎝ 2 ⎠
(b) Using the work energy theorem:
WAB = ΔK = K B − K A , i.e., K B = WAB + K A .
1
∴K B = 12.2 J + (5.00 kg )(1.00 m/s)2 = 14.7 J
2
1
= (5.00 kg ) v B2 .
2
∴v B =
2 × 14.7 J
= 2.42 m/s.
5.00 kg
POWER
DISCUSSION PROBLEM [6.3]:
Power is the rate at which work is
done or energy is dissipated (i.e.,
F
quickly or slowly). For example,
A
F
running up a flight of stairs quickly requires more power
than walking up the same flight of stairs slowly. At any
M
B
2M
Start
Finish
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?
instant, the instantaneous power is
! !
!
dW F • d s ! d s ! !
P=
=
= F • = F • v (scalar),
dt
dt
dt
so power can be > 0 or < 0 depending on the angle
!
!
between F and v .
Dimensions: Power ⇒
A: Sled A
B: Sled B
C: Ha! It’s a dead-heat!
=
work done [M][L]2
⇒
time
[T]2 [T]
[M][L]2
.
[T]3
Units: J/s ⇒ watts (W)
1000 W ⇒ 1 kW
1 HP ⇒ 746 W
v"
v=0
ℓ
x"
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
µ k = 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?
fk
x
N
mg
(a) Two ways to find the work done by the frictional
force:
(i) Wf = −f k .ℓ = −(µ k mg)ℓ
= −0.45(1200 kg)(9.81 m/s 2 )(25.5 m)
= −1.35 ×105 J.
1
1
(ii) Wf = ΔK = mv 2 − mv "2
2
2
1
= − (1200 kg)(15.0 m/s) 2 = −1.35 × 105 J.
2
(b) Average power dissipated by the frictional force is
Pav =
Wf −1.35 × 105 J
=
t
3.40 s
= −3.97 ×10 4 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).
Energy and work are one thing ... but power is quite
another.
One-dimensional example:
Look ... one gallon of gas has a
certain energy content and so can
do a fixed amount of work
6
~ 120 × 10 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 only
provide enough power to operate a
jet engine for ~ 1 minute!
(~ 2500 HP).
! !
P = F • v = Fx v x = ma x v x
P
i.e.,
ax =
.
mv 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
dv
P = Fx v x = ma x v x = mv x x
dt
=
d ⎛1
⎞ dK
⎜ mv x2 ⎟ =
,
⎠ dt
dt ⎝ 2
i.e., the instantaneous power at some time t is the rate of
change of kinetic energy at that same time.
We have just seen that the power (P) is related to the
change in kinetic energy ( ΔK) and time by:
ΔK
P=
.
Δt
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?
So, at constant power, Δt ∝ ΔK.
1
But ΔK = Mv f 2 as the motorcycle starts from rest.
2
1
• From 0 → 20 mi/h: ΔK1 = × M × 202 = 200M.
2
1
• From 0 → 60 mi/h: ΔK 2 = × M × 602 = 1800M.
2
∴ΔK 2 = 9ΔK1,
i.e., Δt 2 = 9Δt1.
If Δt1 = 1 s then Δt 2 = 9 s.
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.
ΔK
We found earlier earlier P =
,
Δt
ΔK m
i.e., Δt =
=
v f 2 − v i2 .
P
2P
500 kg
=
v f 2 − v i2
3
2 × 7.46 × 10 W
(
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
(
)
(
)
(
)
)
= 3.35 × 10−2 v f 2 − v i2 .
(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) 20 mi/h to 30 mi/h?
(
)
(c) Δt = 3.35 × 10 −2 (13.5 m/s)2 − (9.0 m/s)2 = 3.39 s.
(a) Since the object is moving along the x-axis, the work
done is:
3.00
W= ∫
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 − x 2 ,
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.
0.50
(6.00 + 4.00x − x2 ).dx
3.00
2
3
= ⎡6.00x + 4.00⎛⎝ x 2 ⎞⎠ − x 3 ⎤
⎣
⎦ 0.50
(
) (
)
⎡
1
⎤
= ⎢6(3.00 − 0.50) + 2 3.002 − 0.502 − 3.003 − 0.503 ⎥
⎣
⎦
3
= 23.5 J.
(b) The instantaneous power at any point is
! !
P = F • v.
Since the motion is confined to the x-axis,
P = Fx v x ,
where Fx is the force and v x the instantaneous velocity at
x = 3.00 m. To get v x 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 ).
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
Since the object is initially at rest K(at 0.5 m ) = 0,
1
then W = K(at 3 m) = mv x2
2
i.e., v x =
(
2W
2 × 23.5 J
=
= 3.96 m/s.
m
3.00 kg
∴P = Fx v x
)
in kinetic energy. We will look at two examples.
1. Springs: consider a spring inside a toy gun.
!
F1
!
F2
!
x1
!
x2
= 6.00 + 4.00(3.0 ) − (3.0)2 N × 3.96 m/s
Fx (at 3 m)
= 35.6 W.
You compress the spring by applying two equal and
!
!
opposite forces F1 and F2. 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 = F1 • Δx1 + F2 • Δx 2 = 2FΔx,
!
!
!
!
since F1 = F2 = F and Δx1 = Δx 2 = Δx.
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
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,
spring has stored the work you did as potential energy.
∴ΔK = 0.
When released the spring will transfer the stored potential
But you did work on the system ... if it didn’t increase the
energy to kinetic energy of the ball.
kinetic energy of the object what happened to the work?
2. Raising an object in the Earth’s gravitational field.
2
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
2
1
h
2
1
h
mg
Wyou = mgh
Wgrav = − mgh
1
h
2
1
h
2
mg
Wyou = mgh
Wgrav = − mgh
h
v
1
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.
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 .
2
h
v
1
When the object is released from rest,
1
ΔK = W21 = mgh = mv 2 ,
2
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.
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
When the object is raised, the work done by the
gravitational force on the object is
If the work done in going from point 1 to point 2 is W12
W12 = (−mg)h = −mgh .
When it is released, the work done by the gravitational
force on the object is:
Thus, the net work done by the gravitational force is
mgh
then (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
W21 = +mgh ,
1 −mgh 2
around any closed path, no matter the route.
1
= 0.
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.
What about non-conservative forces ?
b
a
Question 6.12: In a certain region of space, the force on
an electron is
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 .d A→B = mgµ kd A→B.
!
F = cxˆj,
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),
Consider two paths a and b. Since b > a, then Wb > Wa .
what is the work done on the electron by the force
So the work done against the frictional force does depend
during one complete trip around the loop? Is the force a
on the path! Therefore, the frictional force is a non-
conservative or non-conservative force?
conservative force.
NOTE: The work done by a non-conservative force is
non-recoverable (it produces heat, sound, etc.)
y
(0, ℓ)
"
F=0
"
dℓ
"
F
y
(ℓ, ℓ)
"
dℓ
(0, ℓ)
"
dℓ
"
F
"
F = cℓˆj
"
F=0
x
(0, 0)
(ℓ, 0)
"
dℓ
0
"
" "
The force ( F = cxˆj) increases linearly with x, but F⊥dℓ ,
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 xdirection, i.e., W12 = W34 = 0.
Along the leg from (ℓ, 0) to (ℓ, ℓ) the work done is
ℓ"
ℓ
" ℓ
W23 = ∫ F • d ℓ = ∫ cxˆj • dyˆj = cℓ ∫ dy =cℓ2 .
0
(ℓ, ℓ)
"
dℓ
"
F
"
dℓ
ˆj
"
F = cℓˆj
ˆi
x
(0, 0)
Along the leg from (0,0) to (ℓ, 0) the work done is
ℓ"
"
W12 = ∫ F • d ℓ .
0
"
dℓ
"
F
0
"
dℓ
(ℓ,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
0
0
0
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 nonconservative force. Also, since W > 0, the electron has
gained energy as it goes around the loop.
Conservative forces and the potential energy function
Non-conservative forces and the work-energy theorem
two forces to consider; the gravitational force
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
(conservative) and the drag force (non-conservative).
position.
Consider an object falling with air-resistance. There are
1
The total work done is
W12 = Wgrav + Wnc ,
2
One property of a conservative force is that the work done
where Wgrav is the work done by the
by the force can be expressed as the difference between
conservative gravitational force and
the initial and final values of the potential energy, i.e.,
! !
dW = F • d s = −dU,
Wnc is the work done by the nonconservative 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 = E mech2 − E mech1 = ΔE mech .
Therefore, the work done by a non-conservative force is
equal to the change in mechanical energy.
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
2!
!
ΔU = U2 − U1 = − ∫ F • d s .
1
! !
Since from above, dU = − F • d s , then, in one-dimension,
dU
dU = −Fx .dx, i.e., Fx = − .
dx
Given that Fx = −
dU
, if we know the potential energy
dx
function, U(x), we can determine the force Fx at any
Note: U ⇒ U(y), i.e., it has
U(y)
2
1
U!
ΔU = U2 − U1
Height (y)
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:
ˆj
!
mg
Let us find the potential energy
function for the gravitational force.
We take the y-direction ( ˆj)
vertical. Then, from above, the incremental change in
potential energy as the object falls is
! !
dU = − F • d s = −(−mgˆj) • dyˆj = ( mg )dy.
∴U(y) = mg ∫ dy = mgy + U" ,
where U" is the reference energy when y = 0. Let’s plot
this function:
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
dU
potential energy function is Fz = −
= −mg, as
dz
expected.
[2] Elastic force:
F2 = − kx
F1 = kx
If you compress a spring
a distance x in the xdirection, the force, F1,
x
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.
Therefore, the elastic potential energy function U(x) of
the spring is given by:
! !
dU = − F • d s = − ( −F2 ).dx = kx.dx.
1
∴U(x) = k ∫ x.dx = kx 2 + U" .
2
Here are three examples of equilibrium where
U(x)
Equilibrium position: x !.
Stable equilibrium when
U(x ! ) is a minimum,
When there is no displacement, i.e., x = 0, the spring has
x
zero elastic potential energy.
∴U(0) = 0, so U" = 0.
dU
= 0:
dx
d2U
i.e.,
> 0.
dx 2
x!
U(x)
the x-direction); we say that the spring is in equilibrium.
Equilibrium position: x !.
Unstable equilibrium when
This is the definition of equilibrium, which we first came
U(x ! ) is a maximum,
At this point, there is no net force acting on the spring (in
across in chapter 4 (Newton’s 1st Law).
x
Equilibrium: At equilibrium
F = 0, i.e.,
dU
= 0,
dx
d2U
i.e.,
< 0.
dx 2
x!
U(x)
Equilibrium position: x !.
Neutral equilibrium since
so U(x) is an extremum. However, we must exercise care
dU
in using that criterion, since
= 0 does not necessarily
dx
U(x ! ) is an inflexion point,
x
define a minimum in the potential energy function.
x!
i.e.,
d2U
= 0.
dx 2
(a) The force associated with the potential function is
F(x)
Question 6.13: The potential energy function of an
object, of mass 2.5 kg, confined to move along the xaxis, is
F(x) = −
dU(x)
= −6x + 6x 2 .
dx
U(x) = 3x 2 − 2x 3,
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,
x(m)
−1
1
2
(b) At equilibrium F(x) = 0,
i.e., 6x 2 − 6x = 6x(x − 1) = 0
Therefore, the equilibrium positions are at x = 0 and
(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.
x = 1 m. To find the type of equilibrium look at the signs
of the seond derivative,
d 2 U(x)
dx 2
Since U(x) = 3x 2 − 2x 3, then
• At x = 0:
, at x = 0 and x = 1 m.
d 2 U(x)
= 6 − 12x.
dx 2
d 2 U(x)
= 6 (> 0) ⇒ Stable.
dx 2
• At x = 1 m:
d 2 U(x)
= −6 (< 0) ⇒ Unstable.
dx 2
(c)
U(J)
5
U(x) = 3x 2 − 2x 3
x(m)
−1
1
2
−5
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.