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LESSON 6
Work, Energy, and Power

Work done by a Constant Force

Work involves force and displacement. We use this quantitatively when a
force moves an object through a distance. In other words, the work done by a
constant force in moving an object is equal to the product of the magnitudes of
the displacement and the component of the force in the direction of the
displacement.

In symbol, this is expressed as,

W = F Cosθ d
Where: W = work done
F = force
d = distance
θ = angle between the displacement and F

The SI unit of work is Joules (J)

1J = 1 N.m
Work is a scalar quantity.

According to this definition, work is done by force F on an object under the

following conditions: (1) the object must undergo a displacement, and (2) F must have
a nonzero component in the direction of the displacement. A force does no work on an
object if the object does not move. For example, if a person pushes against a wall, a
force is exerted on the wall but the person does no work since the wall is fixed. The
wall has not undergone a displacement. However, the person’s muscles are
contracting (undergoing displacement) in the process so that the internal energy is
being used up.

The sign of work also depends on the direction of the force relative to
displacement. The work done by applied force is positive when the vector associated
with the component of the force is in the same direction as the displacement. However,
when the vector associated with the component of the force is opposite the direction of
the displacement, the work done is negative.

Example 1:
A crate of watermelon is pulled a distance of 4m with a horizontal force of 50N.
A frictional force of 8N retards the motion. Calculate:
(a) the work done by the 50-N force
(b) the work done by the frictional force
(c) the total work done on the box

Solution:
(a) The 50N force is in the same direction as the displacement, so θ = 0o.
WF = F cosθ d = 50N cos 0o (4m) = 200 N-m = 200 J

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(b) The frictional force is opposite the direction of motion (opposite the direction
of displacement), so θ = 180o.
Wf = f cosθ d = 8N (cos 180o) 4m = -32 J

(c) WTOTAL = WF + Wf = 200J + (-32J) = 168 J

Example 2:
A 20 kg box of books is pushed along a rough, horizontal surface by a constant
force of 70N acting at an angle of 15o below the horizontal. The block is displaced a
distance of 3m, and the coefficient of kinetic friction is 0.25. Determine the work done
by:
a) the 70-N force
b) the force of friction
c) the normal force
d) the force of gravity

Solution:

Free-body diagram of the box:

a) WF = F cos θ d = (70N cos 15o ) 3m = 202. 8 J

b) Solving for the normal force,

∑Fv = 0
N – Fy – W = 0
N = 70N sin 15o + 20kg (9.8m/s2) = 214.1 N

Solving for the frictional force,

f = µN = 0.25 (214.1N) = 53.5N

Hence, the work done by the frictional force is,

Wf = f cos θ d = (53.5N cos 180o) (3m) = -160.5 J

c) Work done by the normal force is zero because there is no displacement

parallel to the normal force.

d) There is no work done by gravity because there is no vertical displacement.

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Practice Problems:

1. A horizontal force of 150N is applied to a crate which moves a distance of 8m along

a frictionless horizontal surface. Find the work done by the force.

2. If a construction worker lifts a 15-kg bucket of cement mixture from the ground to
the top of a 6-storey building and does 2 kJ of work, how high is the building?

Assignment:

1. A shopper in a supermarket pushes a cart with a force of 40N directed at an angle

of 30o below the horizontal. The force is just sufficient to overcome various frictional
forces, so the cart moves at constant speed. (a) What is the work done by the shopper
as she moves down a 20-m length aisle? (b) The shopper goes down the next aisle,
pushing horizontally and maintaining the same speed as before. If the work done by
frictional forces doesn’t change, would the shopper’s applied force be larger, smaller,
or the same? What about the work done on the cart by the shopper?

Kinetic Energy and Work-Energy Theorem

Kinetic Energy is the energy a body has by virtue of its motion. A car or a bullet
in motion, or a stream of water possesses kinetic energy. The kinetic energy of the
moving object can be measured by the amount of work it will do if brought to rest.

The kinetic energy of a body of mass m and velocity v is expressed as

KE = ½ mv2

The SI unit of kinetic energy is Joule (J).

1 J = 1 N.m = 1 kg. m2/s2

Consider a body with an initial velocity vi on which a steady unbalanced force

Fnet acts as it moves a distance s. Using Newton’s 2nd Law of Motion,

Fnet = ma

The work done by the unbalanced force on the body is W =Fnet s.

W = Fnet s = (ma) s

Using equation 6 in module 3 (Motion in One Dimension),

(vf )2 = (vi )2 + 2as

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( v f ) 2 − ( vi ) 2
as =
2

We can substitute this expression in W = (ma) s to get

 ( v f ) 2 − ( vi ) 2 
W =m 
 2 
 

W = ½ m(vf )2 – ½ m(vi )2

W = KEf - KEi = ∆KE (This equation is called the work-energy

theorem).

Example 1:
Jason pushes a 3.00 x 103 kg car along a horizontal ground from rest to a
speed v, doing 18kJ of work in the process. During this time, the car moves 15.0 m.
Neglecting friction between the car and the ground, find (a) v and (b) the horizontal
force exerted by Jason on the car.

Solution:
(a) W = ∆KE = KEf - KEi = ½ m(vf )2 - ½ m(vi )2
18,000J = ½ (3.00 x 103kg) v2 - 0
v = 3.46 m/s

(b) W = Fcosθ d
18,000J = (F cos 0o) 15m
F = 1,200 N

Example 2:
Starting from rest, a 15-kg block on a horizontal surface acquires a kinetic
energy due to a constant horizontal force of 300N. If the coefficient of kinetic friction is
0.10, find the kinetic energy acquired after the object has traveled 6m.

Solution:

W = Fnet s = (300N – f ) 6m

Solving for f:
f = µkN = 0.10 (15 kg) 9.8 m/s2 = 14.7 N

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FBD of the block:

Therefore the work done by the net force is

W = (300N – 14.7N ) 6m = 1,711.8 J

Using the work-energy theorem,

W = ∆KE = KEf - KEi
1,711.8 J = KEf - 0
KEf = 1,711.8 J

Practice Problems:

1. An automobile with a mass of 1200kg travels at a speed of 70km/h. (a) What is its
kinetic energy? (b) What is the net work that would be required to bring it to rest?

2. A constant net force of 80N acts on an object initially at rest and acts through a
distance of 2m. (a) What is the final kinetic energy of the object? (b) If the object’s
mass is 0.50kg, what is its final speed?

3. A 1,500kg car is traveling at 60 mi per hr. What average force is required to stop the
car in 65m?

Assignment
1. A 500-g object has a speed of 3 m/s at point A and kinetic energy of 9 J at point B.
(a) What is its kinetic energy at point A? (b) What is its velocity at point B? (c) What is
the total work done on the object as it moves from point A to point B?

2. A 0.4 kg ball has a speed of 14 m/s. (a) What is its kinetic energy? (b) If its speed is
doubled, what is its kinetic energy?

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Gravitational Potential Energy

When an object moves in the presence of the earth’s gravity, the gravitational
force can do work on that object. In the case of freely falling object, the work done by
gravity is a function of the vertical displacement of the object.

Gravitational potential energy is the energy a body has by virtue of its position.
In this case, position refers to the height of an object above some reference point, such
as floor or the ground. Suppose that an object of mass m is lifted a distance ∆h (see
figure below). The gravitational potential energy of the object is

PE = mg∆h = mg (h – ho )

If ho = 0, then

PE = mgh
F

∆h
mg

ho

Always consider a reference mark in solving for gravitational potential energy.

Above a reference mark, potential energy is positive; below it, negative.

The unit of potential energy is Joule.

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Example 1:
A 0.30kg ball hangs from a ceiling at the end of a 1.0-m string. If the ceiling is
3.0m from the floor, determine the gravitational potential energy of the ball relative to
(a) the ceiling (b) the ball (c) the floor.

Solution:

(a) PE = mgh = -(0.30kg)(9.8 m/s2) (1.0m) = -

2.94 J
The ball is below the reference mark (ceiling),
so PE is negative.

(b) PE = mgh = (0.30 kg) (9.8 m/s2) (0 ) = 0 J

(c) PE = mgh = (0.30 kg) (9.8 m/s2) (2 m) = 5.88 J

The ball is above the reference mark (floor).
Hence, PE is positive.

Practice Problems:

1. What is the gravitational potential energy of a 20 kg at the top of a 15-m building?

(Take the ground as the reference mark.)

2. A 0.15-kg stone is held 1.5 m above the top edge of a water well and then dropped
into it. The well has a depth of 6.0m. Taking the top of the water well as the reference
mark, what is the gravitational potential energy of the stone (a) before the stone is
released (b) when it reaches the top edge of the well (c) when it reaches the bottom of
the well.

3. A 20-kg rock is on the edge of a 100m cliff.

a. What is the potential energy of the rock relative to the base of the cliff?
b. The rock falls from the cliff. What is its KE just before it hits the ground?
c. What velocity does the rock have just before it hits the ground?

Assignment:
1. A 50.0 kg shell is shot from a cannon at Earth's surface to a height of 400 m.
a. What is the gravitational potential energy of the shell when it is at this height?
b. What is the change in potential energy of the shell when it falls back down to a
height of 100m off the ground?

2. A physics book, mass unknown, is dropped 4.50 m. What velocity does the book
have just before it hits the ground?

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Conservation of Mechanical Energy

Conservation principles play an important role in physics. When a physical

quantity is conserved, the numeric value of the quantity remains the same throughout
the process. Although the form of the quantity may change in some way, its final value
is the same as its initial value.

The sum of kinetic and potential energies is called the total mechanical energy.

ET = PE + KE

In any isolated system of objects interacting only through conservative forces,

the total mechanical energy of the system is constant at all times, or conserved; that
is:

E1 = E2
KE1 + PE1 = KE2 + PE2

Example 1:
A 0.300-kg ball is thrown vertically upward with an initial speed of 10.0 m/s. If
the initial potential energy is zero, find the ball’s kinetic, potential and mechanical
energies at (a) its initial position (b) at 2.50 m above the initial position (c) at its
maximum height.

Solution:
(a) If the potential energy at the initial point is zero, then, the reference mark is the
initial point.

KEi = ½ m v2 = ½ (0.300 kg) (10.0 m/s )2 = 15 J

PE = 0
ET = KE + PE = 15 J + 0 = 15 J

(b) Using the equation in freely falling bodies,

(vf )2 = (vi )2 + 2gs = (10.0 m/s )2 + 2(-9.8 m/s2) 2.50 m
vf = 7.14 m/s

KE = ½ m v2 = ½ (0.300 kg) (7.14 m/s )2 = 7.65 J

PE = mgh = (0.300 kg ) (9.8 m/s2 ) 2.50 m = 7.35 J
ET = KE + PE = 7.65 J + 7.35 J = 15 J

(c) Velocity at the top of the flight is zero.

Solving for h to get PE,
(vf )2 = (vi )2 + 2gs
0 = (10.0 m/s )2 + 2 (-9.8 m/s2) h
h = 5.10 m

PE = mgh = (0.300 kg ) (9.8 m/s2 ) (5.10 m) = 15J

KE = 0
KE + PE = 15 J + 0 = 15 J

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Example 2:
The bob of a swinging pendulum has a speed of 2m/s when it is at the lowest
point. The bob has a mass of 50g and the cord is 1.0m long. (a) What maximum
height will the bob attain? (b) What angle with the vertical does the cord make when
the bob is at its highest point?

(a) Let us take point 2 as the reference point and apply the law of
conservation of mechanical energy,

PE1 + KE1 = PE2 + KE2

mgh1 + ½ mv12 = mgh2 + ½ mv22
(0.050kg) (9.8m/s2) h + ½ (0.050kg) (0) = (0.050kg) (9.8m/s2) (0) + ½
(0.050kg) (2m/s)2
0.49 h = 0.1
h = 0.204m

b)

PE1 + KE1 = PE2 + KE2

mgh1 + ½ mv12 = mgh2 + ½ mv22
(0.050 kg)(9.8m/s2)(−1.0mcosθ)+0 = (0.050kg)(9.8m/s2)(−1.0m)+ ½ (0.050kg)
(2m/s)2
−0.49 cosθ = −0.49 + 0.1
θ = 37.25o

Practice Problems:
1. The bob of a pendulum weighs 40N and is attached to the end of a 1.0 m long
string. The bob is displaced so that the string makes an angle of 20o with the vertical.
Find the kinetic energy and the speed of the bob at the lowest point of the arc.

2. The brakes of a 2,000kg car are released when the car is going down a hill. If the
car is moving at 2m/s when the breaks are released and neglecting friction, how fast
is the car moving when it is 5 m below the point where its velocity is 2m/s?

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3. A 15 kg object is dropped from a height of 30m. Calculate the kinetic and potential
energies of the body (a) at the initial point (b) at a height of 20m (c) 2 seconds after
release.

4. A stone weighing 5 N is thrown vertically upwards with a velocity of 40.0 m/s. Find
(a) its potential and kinetic energy 1.5 s after it is thrown (b) the maximum height
reached applying conservation of mechanical energy.

5. Tarzan swings on a 25.0-m-long vine initially inclined at an angle of 35o with the
vertical. What is his speed at the bottom of the swing (a) if he starts from rest? (b) if he
pushes off with a speed of 6.0 m/s?

Power

Power is the rate of doing work. It is related to how fast a job is done. Two
identical jobs or tasks can be done at different rates - one slowly or and one rapidly.
The work is the same in each case (since they are identical jobs) but the power is
different. The equation for power shows the importance of time. Its average value in a
given time is obtained by dividing the work done by the corresponding time.

Work
Power =
Time

W
P=
t

The SI unit of power is joule per second.

1 joule/second = 1 watt (W)

1 joule = 1 watt-second

A smaller unit of power is the erg per second. In the British engineering system,
the unit of power is ft-lb/s. A larger unit of power is the horsepower.

1 horsepower (Hp) = 550 ft-lb/s

Another larger unit of power is the kilowatt (kW) which is equal to 1000W.

1 kW x 1 h = 1 kW ∙ h
= 1000 W x 3,600 s
= 3.6 x 106 watt-seconds
or, 1 kilowatt-hour = 3.6 x 106 joules

The kW ∙ h is a convenient unit for domestic purposes. We pay for our

electrical bills according to the number of kilowatt-hours we have consumed.

Another useful conversion is

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1 horsepower = 0.746 kilowatt = 746 watts

which means that 1 Hp is approximately ¾ kilowatt.

Another way to express power is in terms of velocity.

Work F x s s
Power = = = F
t t t

Since v = s / t then power can be expressed as

P = Fv

Example 1:
An elevator of mass 2,000 kg rises 40 m in 10 seconds. What is the power
developed?

Solution:
P = W / t = (2,000kg )(9.8 m/s2) 40m / 10 s
P = 78,400 W

Example 2:
The electric motor of a toy train accelerates the train from rest to 0.600 m/s in
15.0 ms. The total mass of the toy train is 525 g. Find the average power delivered to
the train during this acceleration.

Solution:
W
P =
t

Using the work-energy theorem to solve for the work done,

W = ∆KE = KEf - KEi

W = ½ m (vf )2 - ½ m (vi )2

W = ½ (0.525 kg ) (0.600 m/s )2 - 0

W = 0.095J
W
Substitute this value to P = ,
t
0.095 J
P = = 6.33 W
15.0 x 10 − 3 s

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Practice Problems:

1. A tractor pulls a wagon with a constant force of 600 N at a constant speed of 15

km/h. (a) How much work is done by the tractor in 4.0 minutes? (b) What is the
tractor’s power output?

2. A 70-kg soldier in basic training climbs a 10-m vertical rope at uniform speed in 9
seconds. What is his power output?

3. A weightlifter lifts 250 kg through 2 m in 1.5 s. What is his power output?

4. A 65-kg athlete runs a distance of 500 m up a mountain inclined 20o to the

horizontal. He performs this run in 90 seconds. Assuming air resistance is negligible,
(a) how much work does he perform? (b) What is his power output during the run?

5. A 24,000 kg rocket takes off from its launching pad and acquires a vertical velocity
of 14,400 km/h when it is at an altitude of 30km. (a) Find its potential and kinetic
energies. (b) If it reached this altitude in 15 minutes, what is the power developed?

M.E. Omiles Lesson 6. Work, Energy and Power Physics for Engineers