Chapter 7
Kinetic Energy and Work
Kinetic Energy
Definition: The Energy (E) of a body is its capacity to do work.
Energy is present in the Universe in various forms
including mechanical, electromagnetic, chemical,
thermal and nuclear.
The unit of energy is the Joule (J). There are three types
of mechanical energy. We are concerned with two of
them.
Kinetic Energy

Energy is required for any sort of motion

Energy:
o
Is a scalar quantity assigned to an object or a system of objects
o
Can be changed from one form to another
o
Is conservedin a closed system, that is the total amount of energy of all types is
always the same.
Definition: The Kinetic Energy (𝐸𝐾 ) or The energy of
motion : is the work done on a body inbringing it to rest.
1
𝐸𝐾 = 𝑚. 𝑣 2 = 𝐾𝐸
2
𝐸𝐾 : Kinetic Energy (J)
𝑚 : Mass (kg)
𝑚
𝑣 : Velocity ( , 𝑚. 𝑠 −1 )
𝑠
Kinetic Energy

Kinetic energy:
o
The faster an object moves, the greater its kinetic energy
Kinetic energy is zero for a stationary object
For an object with v well below the speed of light:
o

Eq. (7-1)

The unit of kinetic energy is a joule (J)
Eq. (7-2)
Example 1: A 5 kg object moving at a speed 6 m/s. Calculate its kinetic energy.
Solution:
The kinetic energy is:
1
1
𝐾 = 𝑚𝑣 2 = × 5 × 62 = 0.5 × 5 × 36 = 90 𝐽
2
2
Kinetic Energy
Example Energy released by 2 colliding trains with given
weight and acceleration from rest:
o
Find the final velocity of each locomotive:
o
Convert weight to mass:
o
Find the kinetic energy:
Kinetic Energy
Example 2:A car of mass 1000 kg travelling at 30m/s has its speed reduced to 10m/s by a
constant breaking force over a distance of
75m.
Find:
a) The cars initial kinetic energy .
b) The final kinetic energy .
c) The breaking force .
Solution :
1
1
a. Initial kinetic energy = 2 𝑚𝑣 2 = 2 × 1000 × 302 = 500 × 900 = 450000 𝐽 = 450 𝐾𝐽
1
1
b. Final kinetic energy = 2 𝑚𝑣 2 = 2 × 1000 × 102 = 500 × 100 = 50000 𝐽 = 50 𝐾𝐽
c. Change in kinetic energy = 450 − 50 = 400 KJ
𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = Change in kinetic energy
𝐹. 𝑠 = 400000
400000
Breaking force × 75 = 400000 ∴ Breaking force =
= 5333 𝑁
75
Work
Definition: Work (W) is the force multiplied by the distance moved
by the point of application of the force.
SI Unit : 𝑁𝑒𝑤𝑡𝑜𝑛 . 𝑚𝑒𝑡𝑒𝑟 = 𝐽𝑜𝑢𝑙𝑒 , 𝑁 . 𝑚 = 𝐽
Work can be positive, negative, or zero.
The sign of the work depends on the direction of
the force relative to the displacement.
•Work positive : W > 0 if 90 °> θ > 0
•Work negative : W < 0 if 180° > θ > 90°
•Work zero : W = 0 if θ = 90°
•Work maximum if θ = 0°
•Work minimum if θ = 180°
Work and Kinetic Energy


Account for changes in kinetic energy by saying energy has
been transferred to or from the object
In a transfer of energy via a force, work is:
o

Done on the object by the force
This is not the common meaning of the word “work”
o
To do work on an object, energy must be transferred
o
Throwing a baseball does work
o
Pushing an immovable wall does not do work
Work and Kinetic Energy

For an angle φ between force and displacement:
Eq. (7-7)

As vectors we can write:
Eq. (7-8)

Notes on these equations:
o
Force is constant
o
Object is particle-like (rigid)
o
Work can be positive or negative
Work and Kinetic Energy
Figure 7-2

Work has the SI unit of joules (J), the same as energy

In the British system, the unit is foot-pound (ft lb)
Work and Kinetic Energy


For two or more forces, the net work is the sum of the
works done by all the individual forces
Two methods to calculate net work:


We can find all the works and sum the individual work terms.
We can take the vector sum of forces (Fnet) and calculate the net
work once
Work and Kinetic Energy

The work-kinetic energy theorem states:
Eq. (7-10)

(change in kinetic energy) = (the net work done)

Or we can write it as:
Eq. (7-11)

(final KE) = (initial KE) + (net work)
Work and Kinetic Energy

The work-kinetic energy theorem holds for positive and
negative work
Example If the kinetic energy of a particle is initially 5 J:
o
A net transfer of 2 J to the particle (positive work)
• Final KE = 7 J
o
A net transfer of 2 J from the particle (negative work)
• Final KE = 3 J
Work and Kinetic Energy
Answer: (a) energy decreases (b) energy remains the same
(c) work is negative for (a), and work is zero for (b)
Work Done
Example 1: Find the work done on a body in moving it 3 m using a force of 5
N.
Solution :
𝑊 = 𝐹. 𝑠 = 5 × 3 = 15 𝐽
Example 2: A 500 kg body is lifted up )‫ (رفعت‬through 20 m. Find the work
done. ( g =
𝑚
9⋅8 2)
𝑠
.
Solution :
𝑊 = 𝐹. 𝑠 = 𝑚𝑔ℎ = 500 × 9.8 × 20 = 98,000 𝐽
Example 3: A student pulls a vacuum cleaner with a force of 40 N at an angle of
28o to the horizontal. The vacuum cleaner moves 2 m to the right. Find the work
done.
Solution :
𝑊 = 𝐹. 𝑠 cos 𝜃 ⇒ 𝑊 = 40 × 2 × cos 28 = 70.6 𝐽
Work Done by the Gravitational Force

We calculate the work as we would for any force

Our equation is:
Eq. (7-12)

For a rising object:
Eq. (7-13)

For a falling object:
Eq. (7-14)
Work Done by the Gravitational Force

Work done in lifting or lowering an object, applying an
upwards force:
Eq. (7-15)

For a stationary object:
o
Kinetic energies are zero
o
We find:
Eq. (7-16)

In other words, for an applied lifting force:
Eq. (7-17)

Applies regardless of path
Work Done by the Gravitational Force



Figure 7-7 shows the orientations
of forces and their associated
works
for
upward
and
downward displacement
Note that the works (in 7-16)
need not be equal, they are only
equal if the initial and final
kinetic energies are equal
If the works are unequal, you will
need to know the difference
between initial and final kinetic
energy to solve for the work
Figure 7-7
Work Done by the Gravitational Force
Examples You are a passenger:
o
Being pulled up a ski-slope
• Tension does positive work, gravity does negative work
Figure 7-8
Example 1:A horizontal force F is applied to move a 6 kg carton across the floor. If the acceleration of
𝒎
𝒔
the carton is measured to be 𝟒 𝟐 , how much work dose F do in moving the carton 8 m?
Solution:
𝑾 = 𝑭𝒅𝑭 =?
𝑭 = 𝒎𝒂 = 𝟔 × 𝟒 = 𝟐𝟒 𝑵
⟹ 𝑊 = 𝐹𝑑 = 24 × 8 = 192 𝐽
Example 2:A 5 kg box is raised a distance of 2.5 m from rest by a vertical applied force of 90 N. Find:
a) The work done on the box by the applied force.
b) The work done on the box by gravity.
Solution:
a) 𝑊𝐹 = 𝐹𝑑 = 90 × 2.5 = 225 𝐽
b) 𝑊𝐺 = −𝑚𝑔𝑑 = −5 × 9.8 × 2.5 = −122.5 𝐽
Work Done by a Spring Force



A spring force is the variable force from a spring
o
A spring force has a particular mathematical form
o
Many forces in nature have this form
Figure (a) shows the spring in
its relaxed state: since it is
neither
compressed
nor
extended, no force is applied
If we stretch or extend the
spring it resists, and exerts a
restoring force that attempts to
return the spring to its relaxed
state
Figure 7-10
Force of a Spring:
If the spring is either stretched or compressed a small distance from its
equilibrium position, it exerts on the block a force:
𝐹𝑠 = −𝑘𝑥
Wherexis the position of the block relative to its equilibrium (x=0) position
and 𝒌is a positive constant called the spring constant of the spring.
Work Done by a Spring Force

The spring force is given by Hooke's law:
Eq. (7-20)




The negative sign represents that the force always opposes
the displacement
The spring constant k is a is a measure of the stiffness of
the spring
This is a variable force (function of position) and it exhibits
a linear relationship between F and d
For a spring along the x-axis we can write:
Eq. (7-21)
Work Done by a Spring Force
Answer: (a) positive
(b) negative
(c) zero
Power


Power is the time rate at which a force does work
A force does W work in a time Δt; the average power due to
the force is:
Eq. (7-42)

The instantaneous power at a particular time is:
Eq. (7-43)


The SI unit for power is the watt (W): 1 W = 1 J/s
Therefore work-energy can be written as (power) x (time)
e.g. kWh, the kilowatt-hour
Power

Solve for the instantaneous power using the definition of
work:
Eq. (7-47)

Or:
Answer: zero (consider P = Fv cos ɸ, and note that ɸ = 90°)
Eq. (7-48)
Power
Example 1:A 1200 W hair dryer ‫الشعر المجفف‬is on for 8 minutes. How many Joules of energy are
used up?
Solution:
P = 1200 W
t = 8×60 = 480 s
Work (Energy) W = ?
𝑃=
𝑊
∴ 𝑊 = 𝑃 × 𝑡 = 1200 × 480 = 5.76 × 105 𝐽
𝑡
Example 2: A constant force of 2kN pulls ‫ تشد‬، ‫ تسحب‬a crate ‫ صندوق‬along a level floor a distance of
10 m in 50s.What is the power used?
Solation:
𝑊 = 𝐹. 𝑠 ⇒ 𝑊 = 2000 × 10 = 20000 𝐽
𝑃=
𝑊 20000
=
= 400 𝑊
𝑡
50
Power
Example 3:Ahmed dose 300 J of work in pushing a box. If he does the work in 6 s, what is his power
output?
Solution:
𝑊 300
=
= 50 W
𝑡
6
Example 4:Mohammed pushes a box along a smooth floor using a force of 210 N and a power output of
350 W. How long does it take Mohammed to push the box 20 m?
𝑃=
Solution:
𝑊
𝑊
⟹ 𝑡 = 𝑊 =?
𝑡
𝑃
𝑊 = 𝐹𝑑 = 210 × 20 = 4200 𝐽
𝑃=
𝑊 4200
⟹𝑡=
=
= 12 𝑠
𝑃
350
Example 5: A box is lifted at a constant speed of 7.0 m/s by a machine. If the power of the machine is
21000 W, find the force applied.
Solution:
Summary
Kinetic Energy

Work
The energy associated with
motion
Eq. (7-1)
Work Done by a Constant
Force
Eq. (7-7)
Eq. (7-8)

The net work is the sum of
individual works

Energy transferred to or from an
object via a force

Can be positive or negative
Work and Kinetic Energy
Eq. (7-10)
Eq. (7-11)
Summary
Work Done by the
Gravitational Force
Work Done in Lifting and
Lowering an Object
Eq. (7-12)
Eq. (7-16)
Spring Force
Spring Force

Relaxed state: applies no force

Spring constant k measures
stiffness
Eq. (7-20)

For an initial position x = 0:
Eq. (7-26)
Summary
Work Done by a Variable
Force

Found by integrating the
constant-force work equation
Power

The rate at which a force does
work on an object

Average power:
Eq. (7-42)
Eq. (7-32)

Instantaneous power:
Eq. (7-43)

For a force acting on a moving
object:
Eq. (7-47)
Eq. (7-48)