D2.6 Investigate and solve power and work relationships

Ontario Grade 11 Physics

Energy and Society

SPH3U D2.6 — investigating and solving work and power relationships

A force can transfer energy when it causes an object to move. In physics, this transfer is called work. Power describes how quickly work is done. This lesson uses course-level algebra and, when needed, basic trigonometry. First, identify the object or objects being studied. This is the system. Describe motion relative to a reference frame, such as the floor of a room. Choose a positive direction, such as upward or to the right, before assigning signs to forces or displacement. Force and displacement are vectors: each has a size and a direction. Work and power are scalars: each has a size but no direction.

What you will learn

Distinguish work and power, and identify their SI units.
Calculate work using force, displacement, and the angle between them.
Calculate average power from work and time.
Describe a proposed investigation that collects evidence about work and power.

1. From force and displacement to work

A force is a push or pull. Displacement is the change in an object's position, including its direction. Distance tells how much ground an object covers, but displacement also depends on its starting and ending positions. In work calculations, use displacement, not simply the total distance travelled.

In this lesson, positive work means that a force transfers energy to the system. Negative work means that the force transfers energy out of the system. Zero work means that the force transfers no energy through that displacement. The sign depends on the force and displacement directions, not on whether the object is speeding up.

Only the part of a force that points along the displacement contributes to work. If the force points in the same direction as displacement, work is positive. If it points in the opposite direction, work is negative. If it is at a right angle to displacement, work is zero. The angle in the equation is the angle between the force and displacement vectors.

The SI unit of work is the joule, symbol

\mathrm{J}

. One joule is one newton-metre. Work is a scalar, so its answer has no vector direction. Its sign indicates the kind of energy transfer for the chosen system.

W=Fd\cos\theta

Use displacement and the angle between force and displacement.
Work is positive, negative, or zero; it is not a vector.
The SI unit is the joule: $1\ \mathrm{J}=1\ \mathrm{N\,m}$ .

2. Power: how quickly work is done

Power is the rate at which work is done. Average power compares the total work during a time interval with the length of that interval. The SI unit is the watt, symbol

\mathrm{W}

. One watt is one joule per second.

Use the signed work in the relationship. If work is negative, the corresponding average power is negative for the same system and time interval. When a question asks for the amount or magnitude of power, report a positive size and make clear that it is a magnitude.

A fair comparison of power needs both work and time. Doing the same work in less time requires greater average power. Do not confuse power with force: a large force alone does not tell you how much work is done or how quickly it is done.

P_{\mathrm{avg}}=\frac{W}{\Delta t}

Average power depends on both work and elapsed time.
Keep the sign of work when calculating signed average power.
The SI unit is the watt: $1\ \mathrm{W}=1\ \mathrm{J/s}$ .

3. Investigating work and power

An investigation should collect measured evidence rather than assume a result. One proposed classroom procedure is to pull a suitable object along a level surface with a force meter, while measuring the displacement and elapsed time. Record the force reading, the object's displacement in the direction of the pull, and the time interval. Repeat trials if practical, and record the results with units.

For each trial, use the measured force component along the displacement to calculate work. Then divide that work by the measured time interval to calculate average power. A force meter and a timer have limited precision, so record readings as displayed and avoid reporting calculated answers with unjustified extra digits. These are proposed steps; no measurements or outcomes are supplied here.

To investigate how time affects power, compare trials with similar work but different measured times. To investigate how force or displacement affects work, compare trials where the other relevant conditions are kept as similar as practical. State which quantities were measured and which were calculated. A calculated work value is not itself a direct instrument reading.

W=Fd\cos\theta,\qquad P_{\mathrm{avg}}=\frac{W}{\Delta t}

Measure force, displacement, and time; label each with its SI unit.
Distinguish instrument readings from work and power calculated from those readings.
Compare trials carefully and do not describe proposed data as experimental results.

4. A reliable solution method

Start by naming the system and reference frame. Choose a positive direction, then state the known values and the unknown. Draw a quick vector sketch when directions are not obvious. Identify the force whose work is being calculated; several forces can each do separate work on the same object.

Select the work or power relationship before substituting numbers. Convert values to SI units if needed. Keep units in the substitution, use a sensible number of significant figures, and include the sign where relevant. Finish by checking that the units match joules or watts and that the sign and size make physical sense.

1\ \mathrm{J}=1\ \mathrm{N\,m},\qquad 1\ \mathrm{W}=1\ \mathrm{J/s}

Work and power can be calculated for a particular force acting on the chosen system.
Check the angle and direction before deciding the sign of work.
A unit check and reasonableness check are part of the solution.

Worked example

Work by a horizontal pull

A student pulls a cart horizontally with a force of

35\ \mathrm{N}

over a displacement of

4.2\ \mathrm{m}

. Find the work done by the pull.

Set the system and direction
The system is the cart, viewed relative to the floor. Choose right as positive. The pull and the cart's displacement are both to the right, so the angle between them is zero. The unknown is work done by the pull.
Apply the work relationship
Because the force is along the displacement, use the force, displacement, and their angle in the work relationship.
$W=Fd\cos\theta$
Substitute and calculate
Use the given SI values. The cosine of zero is one, so the work is positive.
$W=(35\ \mathrm{N})(4.2\ \mathrm{m})\cos(0^\circ)=147\ \mathrm{J}$

Answer: To two significant figures, the pull does

1.5\times10^2\ \mathrm{J}

of work on the cart.

Check: The units are newton-metres, or joules. Positive work is reasonable because the pull points along the displacement. A few hundred joules is consistent with a force of a few tens of newtons acting over a few metres.

Worked example

Work and average power while lifting

A person lifts a box upward at a steady rate. The upward lifting force is

120\ \mathrm{N}

, the box moves upward

2.5\ \mathrm{m}

, and the lift takes

5.0\ \mathrm{s}

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

Define the situation
The system is the box, relative to the floor. Choose upward as positive. The lifting force and displacement are upward. Find the work done by that force, then divide by the elapsed time to find its average power.
Calculate work
The angle between the lifting force and displacement is zero, so the work is positive.
$W=(120\ \mathrm{N})(2.5\ \mathrm{m})\cos(0^\circ)=300\ \mathrm{J}$
Calculate average power
Average power is the work divided by the time interval. The supplied values support two significant figures.
$P_{\mathrm{avg}}=\frac{300\ \mathrm{J}}{5.0\ \mathrm{s}}=60.\ \mathrm{W}$

Answer: The lifting force does

3.0\times10^2\ \mathrm{J}

of work, at an average power of

6.0\times10^1\ \mathrm{W}

.

Check: The work unit is joules, and joules divided by seconds give watts. Both results are positive because the force and displacement point upward. Doing

300\ \mathrm{J}

of work in

5.0\ \mathrm{s}

gives a reasonable rate of

60.\ \mathrm{J/s}

.

Worked example

Work by a force at an angle

A person pulls a sled with a

50.\ \mathrm{N}

force directed

30^\circ

above the horizontal. The sled moves

6.0\ \mathrm{m}

horizontally to the right. Find the work done by the pull.

Set the direction and angle
The system is the sled, relative to the ground. Choose right as positive. The displacement is horizontal, and the pull is angled upward from that direction. The angle between the force and displacement is $30^\circ$ .
Use the component along the motion
Only the part of the pulling force along the horizontal displacement contributes to work. The cosine factor accounts for that part.
$W=Fd\cos\theta$
Substitute and round
Substitute the SI values. Round the result to two significant figures, matching the least precise supplied value.
$W=(50.\ \mathrm{N})(6.0\ \mathrm{m})\cos(30^\circ)=2.6\times10^2\ \mathrm{J}$

Answer: The pull does

2.6\times10^2\ \mathrm{J}

of work on the sled.

Check: The answer is in joules. It is positive because the pull has a component in the direction of motion. It is less than

300\ \mathrm{J}

, the value for a

50.\ \mathrm{N}

force pointing fully along the

6.0\ \mathrm{m}

displacement, as expected.

Common mistakes and how to avoid them

Using the full force in the direction of motion when the force is angled.

Correction: Use the angle between force and displacement in

W=Fd\cos\theta

.

Treating work as a vector with a direction such as left or upward.

Correction: Work is a scalar. Report its sign and explain what that sign means for the chosen system.

Using distance travelled instead of displacement when direction changes.

Correction: Use the displacement from start to finish and consider the force direction relative to it.

Assuming that a large force always means large power.

Correction: Power depends on work and time. Compare both quantities before making a conclusion.

Reporting work in newtons or power in joules.

Correction: Work is measured in joules. Power is measured in watts, or joules per second.

Lesson summary

Work measures energy transfer by a force during displacement.
The work relationship is $W=Fd\cos\theta$ ; positive, negative, or zero depends on the force-displacement angle.
Average power is work divided by elapsed time: $P_{\mathrm{avg}}=W/\Delta t$ .
Use SI units, significant figures, a stated direction convention, and checks of units and reasonableness.
In an investigation, label which values are measured and which are calculated.

Check your understanding

Question 1

A force points at right angles to an object's displacement. What work does that force do?

Zero work
Positive work equal to force times displacement
Negative work equal to force times displacement
correctIndex: 0

Show answer and explanation

Zero work

At a right angle,

\cos(90^\circ)=0

, so the force does zero work through that displacement.

Question 2

A machine does

240\ \mathrm{J}

of work in

8.0\ \mathrm{s}

. What is its average power?

$30\ \mathrm{W}$
$1.9\times10^3\ \mathrm{W}$
$0.033\ \mathrm{W}$
correctIndex: 0

Show answer and explanation

30\ \mathrm{W}

Average power is work divided by time:

240\ \mathrm{J}/8.0\ \mathrm{s}=30\ \mathrm{W}

. The units reduce to joules per second.

Question 3

A friction force points opposite an object's displacement. What is the sign of the work done by friction?

Negative
Positive
Zero in every case
correctIndex: 0

Show answer and explanation

Negative

The angle between friction and displacement is

180^\circ

, so the cosine is negative and the work done by friction is negative.

Key terms

System: The object or group of objects chosen for study.
Reference frame: The viewpoint used to describe position and motion, such as the floor of a room.
Displacement: The change in position from an object's starting point to its ending point, including direction.
Work: Energy transferred by a force acting through displacement.
Power: The rate at which work is done.
Scalar: A quantity with size but no direction.
Vector: A quantity with both size and direction.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation D2.6. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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D2.6 · Investigate and solve power and work relationships

Ontario Grade 11 Physics

SPH3U D2.6 — investigating and solving work and power relationships

What you will learn

1. From force and displacement to work

2. Power: how quickly work is done

3. Investigating work and power

4. A reliable solution method

Work by a horizontal pull

Work and average power while lifting

Work by a force at an angle

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH3U

About this lesson and its review