Work is done when a force causes an object to move through a displacement. Work can change the energy of a system based on the work-energy theorem, which states that work done on a system equals the change in its energy. Power is the rate at which work is done or energy is transferred, measured in watts. It is calculated by dividing work by time or by multiplying force, velocity, and the cosine of the angle between them.

1. Work, Energy, and Power

2.  Work is done on a system when a force is applied
through a displacement.
 Work is measured in joules.
 One joule of work is done when a force of 1N acts on
a system over a displacement of 1m .
Work

3. Work

4.  Consider a force is exerted on an object while
the object moves a certain distance.
 There is a net force, so the object is
accelerated, a =
𝑓
𝑚
, and its velocity changes.
Work done by a constant force

5.  Recall from the equations of motion
𝑣 𝑓
2
= 𝑣𝑖
2
+ 2𝑎𝑑
 This can be written as𝑣 𝑓
2
− 𝑣𝑖
2
= 2𝑎𝑑
 If you replace (a) with (
𝑓
𝑚
) you obtain :
𝑣 𝑓
2
− 𝑣𝑖
2
= 2(
𝑓
𝑚
)𝑑 multiplying both sides by
𝑚
2
1
2 𝑚 𝑣 𝑓
2
− 1
2 𝑚 𝑣𝑖
2
= 𝑓𝑑
𝐾𝐸𝑓 − 𝐾𝐸𝑖 = 𝑓𝑑
𝑊 = 𝑓𝑑
Work done by a constant force

6. Work done by a constant force
exerted at an angle
𝐹𝑥 = 𝐹 cos 𝜃 = 125𝑁 cos 25° = 113 𝑁.
𝐹𝑥 = 𝐹 sin 𝜃 = 125𝑁 sin 25° = −52.8 𝑁.
 The negative sign shows that the force is
downward.
 Because the displacement is in the x
direction, only the x-component does work.
The y-component does no work.
 The work you do when you exert a force
on a system at an angle to the direction of
motion is equal to the component of the
direction of the displacement multiplied
by the displacement.

7.  Work is equal to the product of the magnitude of the
force and magnitude of displacement times the
cosine of the angle between them.
𝑊 = 𝐹𝑑 cos 𝜃
W : work
F : force
d : displacement
𝜃 : the angle between the two directions of both force
and displacement
Work

8.  When several forces are exerted on a system,
calculate the work done by each force, and then add
the results.
 Suppose you are pushing a box on a friction less
surface while your friend is trying to prevent you from
moving it, as shown in the following figure.
Work done by many forces

9. Work done by many forces

10.  What are the forces are acting on the box?
You are exerting a force to the right and your friend is
exerting a force to the left.
Earth’s gravity exerts a downward force, and the
ground exerts an upward normal force.
Work done by many forces

11.  How much work is done on the box?
The upward and downward forces are perpendicular 𝜃 = 90° to the
direction of motion and do no work. For these forces, 𝜃 = 90°, which
makes cos 𝜃 = 0, and thus 𝑊 = 0.
The force you exert is in the direction of the displacement, so the
work you do is W = 𝐹𝑜𝑛 𝑏𝑜𝑥 𝑏𝑦 𝑦𝑜𝑢 × 𝑑
Your friend exerts a force in the direction opposite the displacement
𝜃 = 180° . Because cos 180° = −1, your friend does negative work.
W = −𝐹𝑜𝑛 𝑏𝑜𝑥 𝑏𝑦 𝑓𝑟𝑖𝑒𝑛𝑑 × 𝑑
The total work done on the box would be
W = (𝐹𝑜𝑛 𝑏𝑜𝑥 𝑏𝑦 𝑦𝑜𝑢× 𝑑) −(𝐹𝑜𝑛 𝑏𝑜𝑥 𝑏𝑦 𝑓𝑟𝑖𝑒𝑛𝑑 × 𝑑)
Work done by many forces

12.  In the last example, the force changed, but we could
determine the done in each segment. But what if the
force changes in a more complicated way?
 A graph of force versus displacement lets you
determine the work done by a force.
 This graphical method can be used to solve problems
in which the force is changing.
Finding work done when forces
change

13. Finding work done when forces
change

14.  The left panel of this figure shows the work done by a constant force of 20.0
N that is exerted to lift an object 1.50 m.
 The work is done by this force is represented by 𝑊 = 𝐹𝑑 =
20𝑁 1.5𝑚 = 30𝐽
 Notice that the shaded area under the left graph is also equal to
20𝑁 1.5𝑚 , 𝑜𝑟 30𝐽
 The area under a force-displacement graph is equal to the force done by
that force.
Finding work done when forces
change

15.  The right panel of this figure shows the force exerted by a
spring, which varies linearly from 0 to 20 N as it
compressed 1.5 m. the work done by the force that
compressed the spring is the area under the graph, which
is the area of a triangle, 𝐴𝑟𝑒𝑎∆ = 1
2
𝑏𝑎𝑠𝑒 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒 , or
𝑊 = 1
2
20𝑁 1.5𝑚 = 15 𝐽
Finding work done when forces
change

16. Use the problem-solving strategies
below when you solve

21.  Energy is the ability of a system to produce a change
in it self or the world around it.
Energy

22.  Work-Energy theorem states that work done on a
system is equal to the change in the system’s energy.
𝑤 = ∆𝐸
 Energy is measured in joules.
Work-Energy Theorem

23.  Through the process of doing work, energy can move
between the external world and the system.
 The direction of energy transfer can be either way.
 if the external world does work on a system, then W
is positive and the energy of the system increases.
 If a system does work on the external world, then W
is negative and the energy of the system decreases.
 Work is the transfer of energy that occurs when a
force is applied through a displacement.
Work-Energy Theorem

24.  Kinetic energy (KE) is the energy associated with
motion.
 Translational kinetic energy (𝐾𝐸𝑡𝑟𝑎𝑛𝑠) is the energy
due to changing position.
𝐾𝐸𝑡𝑟𝑎𝑛𝑠 = 1
2 𝑚𝑣2
 A system’s translational kinetic energy is equal to
1
2
times the system’s mass multiplied by the system’s
speed squared.
Changing kinetic energy

25.  Suppose you had a stack of books to move from the
floor to a shelf.
 You could lift the entire stack at once, or you could
move the books one at a time.
 How would the amount of work compare between
the two cases? Since the total force applied and the
displacement are the same in both cases, the work is
the same.
Power

26.  The time needed is different. Recall that work causes
a change in energy. The rate at which energy is
transformed is power.
 Power is equal to the change in energy divided by the
time required for the change.
𝑃 =
∆𝐸
𝑡
Power

27.  When work causes the change in energy, power is
equal to the work done divided by the time taken to
do the work.
𝑃 =
𝑊
𝑡
 Power is measured in watts (W).
 One watt is 1 J of energy transformed in 1 s.
 1,000 W = 1 KW 1,000,000 W = 1 MW
Power

30. You might have noticed in Example problem 3 that
when the force has a component 𝐹𝑥 in the same
direction as the displacement, 𝑃 =
𝐹𝑥 𝑑
𝑡
.
However, because 𝑣 =
𝑑
𝑡
, power also can be calculated
using 𝑃 = 𝐹𝑣
Power and speed

31. Power and speed