Explain relationships among energy, work, power, and their units

Ontario Grade 11 Physics

Energy and Society

How energy transfer is measured and how quickly it happens

A moving object can gain or lose energy when a force acts on it. In physics, work describes this energy transfer. Power tells us how quickly the transfer happens. These words have precise meanings in physics, and each has a unit that helps us describe and compare situations.

What you will learn

Explain how work describes energy transferred by a force.
Connect the joule, newton, and metre through the definition of work.
Explain how power compares work or energy transfer with time.
Use SI units and check whether a calculated result makes sense.

1. Prerequisite bridge: quantities and units

A scalar quantity has a size but no direction. Energy, work, and power are scalars. A vector quantity has both size and direction. Force and displacement are vectors. Displacement means the change in an object's position, including a direction.

The system is the object or group of objects we are studying. In each example, name the system before calculating. Choose a positive direction to keep track of force and displacement directions. For the basic calculations in this lesson, the force and displacement point in the same direction, so the work is positive.

The SI unit of force is the newton, written

\mathrm{N}

. The SI unit of distance or displacement is the metre, written

\mathrm{m}

. Time is measured in seconds, written

\mathrm{s}

. Energy and work are measured in joules, written

\mathrm{J}

, and power is measured in watts, written

\mathrm{W}

. The watt symbol is not the same thing as the letter

W

sometimes used for work.

Energy, work, and power are scalars.
Force and displacement are vectors, so their directions matter.
Use SI units: newtons, metres, seconds, joules, and watts.

2. Work: energy transferred by a force

In everyday speech, work can mean any effort. In physics, work has a narrower meaning: a force transfers energy when it acts on an object while the object is displaced. If a force and displacement point in the same direction, multiply the force magnitude by the displacement magnitude.

In the relationship,

W

means work,

F

means the force doing the work, and

d

means displacement in the force's direction. Work is measured in joules. One joule is the work done by a force of one newton over a displacement of one metre in the same direction. Thus, a joule is equivalent to a newton-metre.

If an object is held still, its displacement is zero, so the force does no work on it in this model, even if holding it feels tiring. If an object moves in the same direction as the force, work is positive. A direction choice helps describe the situation, but work itself is a scalar and has no direction.

W=Fd

Work connects force and displacement when the force acts along the displacement.
The unit relationship is $1\,\mathrm{J}=1\,\mathrm{N}\cdot\mathrm{m}$ .
No displacement means no work by that force in this model.

3. Power: how quickly energy is transferred

Power describes the rate of energy transfer. Here, rate means an amount divided by the time taken. If two people do the same amount of work, the person who does it in less time produces greater power.

The symbol

P

represents power,

W

represents work, and

t

represents elapsed time. Since work is an energy transfer, power can also be described as energy transferred per unit time. The SI unit of power is the watt. One watt is one joule transferred or used each second.

The equations show how the ideas fit together: force and displacement determine work, and work and time determine power. Keep units in each substitution. A result in joules describes an amount of energy transferred; a result in watts describes how quickly that transfer occurs.

P=\frac{W}{t}

Power is work divided by elapsed time.
The equivalent unit relationship is $1\,\mathrm{W}=1\,\mathrm{J}/\mathrm{s}$ .
Greater power means more energy transferred per second, not necessarily more total energy.

4. Put the relationships together

Start by identifying the system and the direction of motion. If the force points along the displacement, find work using force multiplied by displacement. Then, if a time is given, divide the work by that time to find power. If energy transferred is given instead of work, use that amount in the power relationship.

A useful check is to inspect the units. Force multiplied by displacement gives newton-metres, which are joules. Joules divided by seconds give watts. Also ask whether the result is reasonable: doing the same work in less time should give a larger power.

P=\frac{Fd}{t}

For a force in the displacement direction, calculate work first.
Use elapsed time to relate work or energy transferred to power.
Check unit conversions and compare the result with the situation.

Worked example

Work done while moving a crate

A student pushes a crate with a steady horizontal force of

32\,\mathrm{N}

. The crate moves

4.5\,\mathrm{m}

horizontally in the direction of the push. Find the work done by the push.

Set up the situation
The system is the crate. Choose the push direction as positive. The force and displacement are both in that direction, so the work is positive. The unknown is work.
Choose the relationship
For a force acting along the displacement, work equals force multiplied by displacement.
$W=Fd$
Substitute and calculate
Use the given values with their units. The two significant figures in the measurements support an answer to two significant figures.
$W=(32\,\mathrm{N})(4.5\,\mathrm{m})=1.4\times10^2\,\mathrm{J}$

Answer: The push does

1.4\times10^2\,\mathrm{J}

of work on the crate.

Check: The units are

\mathrm{N}\cdot\mathrm{m}=\mathrm{J}

. The work is positive because the push and displacement point in the same direction. A force of a few tens of newtons acting over several metres should transfer energy on the scale of hundreds of joules, so the result is reasonable.

Worked example

Power while lifting a load

A worker lifts a load with an upward force of

180\,\mathrm{N}

through

2.0\,\mathrm{m}

in

4.0\,\mathrm{s}

. Find the work done by the lifting force and the power.

Define directions and unknowns
The system is the load. Choose upward as positive. The upward force and displacement are in the same direction. Find work first, then power.
Calculate the work
The lifting force transfers energy to the load as it moves upward. Multiply the force by the displacement.
$W=Fd=(180\,\mathrm{N})(2.0\,\mathrm{m})=3.6\times10^2\,\mathrm{J}$
Calculate the power
Power is the work divided by the time taken. Keep the seconds in the substitution.
$P=\frac{W}{t}=\frac{3.6\times10^2\,\mathrm{J}}{4.0\,\mathrm{s}}=9.0\times10^1\,\mathrm{W}$

Answer: The work done by the lifting force is

3.6\times10^2\,\mathrm{J}

, and the power is

9.0\times10^1\,\mathrm{W}

.

Check: The work unit is

\mathrm{N}\cdot\mathrm{m}=\mathrm{J}

, and the power unit is

\mathrm{J}/\mathrm{s}=\mathrm{W}

. The direction is upward for both force and displacement, so the work is positive. Transferring

360\,\mathrm{J}

over four seconds gives

90\,\mathrm{J}

each second, which is consistent with

90\,\mathrm{W}

.

Worked example

Comparing equal work done in different times

Two pumps each transfer

720\,\mathrm{J}

of energy. Pump A takes

12\,\mathrm{s}

, while Pump B takes

8.0\,\mathrm{s}

. Find each pump's power and identify which has greater power.

Identify the quantities
For each pump, treat the energy transferred as the work amount. Power is the unknown. The system is the energy-transfer task for each pump; there is no force or displacement information to calculate work.
Calculate Pump A's power
Divide its transferred energy by its elapsed time.
$P_A=\frac{720\,\mathrm{J}}{12\,\mathrm{s}}=60\,\mathrm{W}$
Calculate Pump B's power
Use the same relationship for Pump B. Report two significant figures, consistent with $8.0\,\mathrm{s}$ .
$P_B=\frac{720\,\mathrm{J}}{8.0\,\mathrm{s}}=9.0\times10^1\,\mathrm{W}$

Answer: Pump A has a power of

60\,\mathrm{W}

, and Pump B has a power of

9.0\times10^1\,\mathrm{W}

. Pump B has greater power.

Check: Both units are joules per second, or watts. Each pump transfers the same energy, but Pump B takes less time, so it transfers more energy per second. Its larger power is reasonable.

Common mistakes and how to avoid them

Using the everyday meaning of work and counting effort alone as physics work.

Correction: For this relationship, a force must act while there is displacement in its direction. With zero displacement, that force does zero work in this model.

Treating work and power as the same quantity.

Correction: Work measures an amount of energy transferred in joules. Power measures how quickly it is transferred in watts.

Writing joules as the unit of power.

Correction: Joules measure energy or work. Power uses watts, and one watt equals one joule per second.

Ignoring the direction of force and displacement.

Correction: Check their directions before using the work relationship. The examples use force and displacement in the same direction, giving positive work.

Lesson summary

Work describes energy transferred by a force acting through a displacement.
For force along displacement, work is force multiplied by displacement, and its unit is the joule.
Power is work or energy transferred divided by time, and its unit is the watt.
Check direction, significant figures, and units in every calculation.

Check your understanding

Question 1

A force of

15\,\mathrm{N}

moves an object

3.0\,\mathrm{m}

in the force's direction. What work is done?

$5.0\,\mathrm{J}$
$45\,\mathrm{J}$
$45\,\mathrm{W}$
$18\,\mathrm{J}$

Show answer and explanation

45\,\mathrm{J}

Work is force multiplied by displacement:

(15\,\mathrm{N})(3.0\,\mathrm{m})=45\,\mathrm{J}

. The unit is a joule, not a watt.

Question 2

A device transfers

240\,\mathrm{J}

in

6.0\,\mathrm{s}

. What is its power?

$40\,\mathrm{W}$
$1.4\times10^3\,\mathrm{W}$
$40\,\mathrm{J}$
$1.4\times10^3\,\mathrm{J}$

Show answer and explanation

40\,\mathrm{W}

Power is energy transferred divided by time:

240\,\mathrm{J}/6.0\,\mathrm{s}=40\,\mathrm{W}

.

Question 3

Two tasks transfer the same energy. Task A takes

5.0\,\mathrm{s}

and Task B takes

10\,\mathrm{s}

. Which statement is correct?

Task B has twice the power of Task A.
Both tasks must have the same power.
Task A has twice the power of Task B.
Task A transfers less energy.

Show answer and explanation

Task A has twice the power of Task B.

For equal energy transfers, the task completed in half the time has twice the power. Task A takes half as long as Task B.

Key terms

Energy: A quantity that can be transferred or used to describe changes in a physical system. Its SI unit is the joule.
Work: Energy transferred when a force acts on an object as it is displaced. Its SI unit is the joule.
Power: The amount of work or energy transferred per unit time. Its SI unit is the watt.
Displacement: The change in an object's position, described with a size and direction.
Scalar: A quantity with a size but no direction.
Vector: A quantity with both a size and a direction.
System: The object or group of objects chosen for study.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Physics (SPH3U), expectation D3.2. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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