Lesson 10: Work, Potential, and Kinetic Energy

Work, Potential, and Kinetic Energy

The mechanical work on an object is the amount of mechanical energy transferred to that object by a
force:
W =Δ E

Energy is a concept so fundamental in physics that it is not easily defined in terms of anything more
fundamental. It is easier just to understand energy in terms of its component kinetic and potential
energies.
Kinetic energy is the energy of motion
and potential energy is the energy to, potentially, do something else.

Thermal energy, the energy of the particles of a substance, is partly kinetic energy and partly potential
energy. (The kinetic energy is in the kinetic energy of the atoms. The potential energy is stored in the
deformation of atomic bonds during this motion.)

kg⋅m2
The SI derived unit of work is the Joule: 1 J =1
s2

But there exist other metric units, e.g. 1 kWh = 3.6 x 106 J or 1 cal = 4.184 J

A calorie (with a lowercase c) is defined as the amount of heat necessary to raise the
temperature of 1 g of water by 1oC (at standard atmospheric pressure).

Food energy is measured in kilocalories (or Calories with a capital C).

In the simplest case of an object moving in one direction and a constant force applied parallel to that
direction:
W =F Δ d

where F is the magnitude of the applied force and Dd is the distance travelled. (Note that work is a scalar.)

If I apply a horizontal force of magnitude 40 N to push a box 15 m across a frictionless surface,

the work done on the box is W =F Δ d =( 40 N )( 15 m)=600 J

But if I apply a vertical force of magnitude 40 N on the same box as it slides 15 m across a
frictionless surface, the work done on the box is zero.

Why? Note that the perpendicular force was not increasing the energy of the box.

The only portion of the force that counts towards the work done is
that component of the force in the direction of the motion:

W =( F cos θ) Δ d

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 Lesson 10: Work, Potential, and Kinetic Energy

If there is a frictional force, it will do negative work on the object, i.e. reduce its total mechanical energy.
Note that the angle between the frictional force vector and the motion is 180o and cos 180o = -1.

If I apply a horizontal force of magnitude 40 N to push a box 15 m across a surface, and the
frictional force is 20 N, the work done by friction on the box is:
W =( F cos θ)Δ d =(20 N )(cos 180 o)(15 m)=−300 J

The total change in energy of the box would be 600 J – 300 J = 300 J.

Gravitational Potential Energy

If you lift an object a distance Dh at a constant velocity, the applied force that does the work must be
equal in magnitude but opposite in direction to the weight of the object:

W =F Δ d =m g Δ h

The energy that this work is increasing is the object’s gravitational potential energy W =Δ E g .

Gravitational potential energy itself is often written as:

E g=m g h where h is understood as being relative to some reference point.

(Note that although ground level is often the reference point, there is no designated spot on Earth
that is your reference point with zero height. You must designate a reference point that is
appropriate to the question.)

Practice Question 1

(a) A force of 30 N [right] is applied to a box of weight 10 N to move it 1 m across the floor. The work
done by the 30 N force on the box is:

A. 10 J B. 20 J C. 30 J D. 40 J

(b) A force of 30 N [right] is applied to a box of weight 10 N to move it 1 m across the floor. If there is
also a 10 N [left] frictional force acting on the box, what is the total change in energy of the box?

A. -10 J B. zero C. 10 J D. 20 J

(c) How much work is required to lift a box of weight 10 N from a height of 1 m to a height of 3 m?

A. 10 J B. 20 J C. 30 J D. 40 J

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 Lesson 10: Work, Potential, and Kinetic Energy

In physics, work and power are not the same thing!

W ΔE
Power is the rate at which work is performed or energy is transferred: P= or
Δt Δt
It has units of Joules per second (J/s) or Watts (W).

W FΔd Δd
Note that P= = =F ( )= F v .
Δt Δt Δt

Practice Question 2

(a) A container factory uses a motor to operate a conveyor belt that lifts containers from one floor to
another. To lift 250 1-kg containers a vertical distance of 3.6 m, the motor runs for 45 s. What is the
power output of the motor?

(b) A bacterium spins its helical flagellum like a rotary motor to overcome the drag force that opposes its
motion in order to propel itself through water. If the bacterium is moving at a constant velocity of 80 μm/s
and the drag force is 0.125 μN, what is this motor's power output?

(If the energy released by the decomposition of one ATP molecule is 5 x10-20 J, how many ATP molecules
per second are required?)

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 Lesson 10: Work, Potential, and Kinetic Energy

Practice Question 3
Determine your work done and your power output while walking and
while running up a flight of stairs at constant speed. (Note: use a complete
flight of stairs, from one floor to the next, ignoring any landing. You can
challenge yourself to run up a longer flight of stairs if you can find one.)

Safety tip: Walk and run safely. Make sure that you are wearing running
shoes with good traction, and remove any obstacles from the stairs. Be
careful not to trip and fall. If you cannot collect data safely, ask your
teacher for sample data.

Measure or estimate your mass in kg. (Remember 1 kg = 2.2 lbs.) Measure the height of one stair using a
ruler and multiply by the number of stairs to calculate the height of the flight of stairs. Use a stopwatch
timer to measure how long it takes you to walk up the stairs and how long it takes you to run up the stairs.

(a) Calculate your work done while walking and while running.

(b) Convert your work done to food Calories (1 food Calorie = 4184 J).

(c) Calculate the number of times you would need to walk or run up the stairs to burn off one Big Mac
hamburger (approximately 600 Calories).

(Given that you do not walk up stairs this often during the day, how do you burn
most of the Calories you consume during the day? Hint: you are warm-blooded.)

(d) Calculate your power output while walking and while running.

(e) Convert your power output to horsepower (1 hp = 745.7 W).

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 Lesson 10: Work, Potential, and Kinetic Energy

Kinetic Energy

When a force is applied to accelerate an object from speed v1 to speed v2, the work done on the object can
be written as:
W =F Δ d =m a Δ d

v 2 −v 1 1
Substituting a= and  d =v av  t= v 2 v 1  t into this equation and simplifying yields:
t 2

1 1
W = m v 22− m v12
2 2

Since this work is done to increase the speed of the object, the energy that is increased is the object's
kinetic energy Ek.
1 2
The kinetic energy at any instant is therefore: E k = mv
2
Note that the direction of the speed does not matter!

For example, if each heartbeat pushes 0.050 kg of blood out of the heart with a speed
of 0.25 m/s, the kinetic energy of the blood is:

1 1
E k = mv 2 = (0.050 kg )(0.25 m/ s)2 =1.6×10−3 J
2 2

Practice Question 4

(a) A toy car of mass 5.0 kg is travelling at a speed of 4.0 m/s. 27 J of work
is done to increase the speed of the toy car. Calculate its final speed.
Hint: calculate the initial kinetic energy. Add the work done to find the final kinetic energy. Rearrange the
equation for kinetic energy to find the final speed.

(b) Which requires more work: increasing the speed of an object from 0 to 5 m/s or from 5 m/s to 10 m/s?

A. from 0 to 5 m/s B. from 5 m/s to 10 m/s

C. both require the same amount of work D. It cannot be determined.

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 Lesson 10: Work, Potential, and Kinetic Energy

Practice Question 5
Determine your work done and your power output while
accelerating to a sprint from rest along a straight line path.

Safety tip: Again, make sure that you are wearing running
shoes with good traction, and remove any obstacles from
your path. Be careful not to trip and fall. Ask your teacher
for sample data if you cannot collect data safely.

Use a measuring tape to measure out a straight-line distance of 10.0 m along a sidewalk or other surface.
Mark the beginning and end of the distance with masking tape or chalk. Start at the beginning mark and
use a stopwatch timer to measure how long it takes you to run to the end mark. Do not slow down before
you reach the end mark. You should be running at a full sprint when your cross the line.

(Why it would not be practical to use a distance of longer than 10.0 m for this activity? And why would
you not have wanted to use a distance that was shorter?)

(a) Use your distance and time measurements (and that your initial speed was zero) to calculate your final
speed and the magnitude of your acceleration.

(b) Use your calculated final speed (and your mass) to calculate your work done while accelerating and
convert your work done to food Calories.

(c) Calculate your power output while accelerating and convert your power output to horsepower.

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