D3.5 Describe when a force does mechanical work

Ontario Grade 11 Physics

Energy and Society

SPH3U study topic D3.5: connect force and displacement

In everyday speech, work can mean effort or a task. In physics, mechanical work has a specific meaning: a force does work when it causes an object to move, or has a component that acts along the object's displacement. Holding a heavy bag still may feel tiring, but the upward force from your hand does no mechanical work on the bag while it remains in place. This lesson focuses on how to tell when a particular force does mechanical work.

What you will learn

Describe the conditions that must be met for a force to do mechanical work.
Use the angle between a force and a displacement to decide whether work is positive, negative, or zero.
Calculate work using the force component in the direction of displacement, with correct units and signs.

1. Prerequisite bridge: force, displacement, and direction

A force is a push or pull. Force is a vector: it has both a size, called its magnitude, and a direction. The SI unit of force is the newton, written

\mathrm{N}

. Displacement describes a change in position from start to finish. It is also a vector, and its SI unit is the metre, written

\mathrm{m}

. Displacement is not the same as total distance travelled.

A scalar has magnitude but no direction. Mechanical work is a scalar. It can be positive, negative, or zero, but it does not point north, south, up, or down.

Before deciding whether a force does work, name the system: the object whose motion you are considering. Also choose a reference frame, which is the viewpoint used to describe its motion. In this lesson, use the ground as the reference frame unless stated otherwise. For a simple calculation, choose the direction of displacement as positive. The angle between a force and the displacement tells us how much the force acts along that displacement.

System: the object being considered.
Reference frame: the viewpoint used to describe position and motion.
Force and displacement are vectors; work is a scalar.

2. The rule: compare force with displacement

A force does mechanical work on an object when the object has a displacement and the force has a component along that displacement. The component is the part of the force that points in the displacement direction. If the force is at an angle, only this along-the-displacement part contributes to the work.

For a constant force and a straight displacement, use the work relationship shown below. Here,

W

is work in joules,

F

is force magnitude in newtons,

d

is displacement magnitude in metres, and

\theta

is the angle between the force direction and the displacement direction. One joule is equal to one newton metre.

If the force points in the same direction as displacement, the angle is

0^\circ

and its work is positive. If the force points partly against the displacement, the angle is greater than

90^\circ

and less than

180^\circ

; its work is negative. If the force is perpendicular to displacement, the angle is

90^\circ

and its work is zero. A force also does zero work if there is no displacement.

Work is calculated for a particular force. Several forces can act on the same object, and each force may do a different amount or sign of work. Do not assume that the work by one force describes the work by every force.

W=Fd\cos\theta

Same-direction force and displacement: positive work.
Opposite-direction component: negative work.
Perpendicular force, or no displacement: zero work.
The SI unit of work is the joule, $\mathrm{J}$ , where $1\,\mathrm{J}=1\,\mathrm{N\,m}$ .

3. Reading the direction before calculating

A quick vector sketch can help. Draw the displacement arrow first. Then draw the force arrow from the same starting point. The angle between the arrows is

\theta

. The part of the force arrow that lines up with the displacement determines the sign and size of the work.

For example, a person pulling a sled forward while it moves forward does positive work on the sled. Friction on the sled points backward while the sled moves forward, so friction does negative work. The normal force from level ground points upward while the sled moves horizontally, so that force is perpendicular to the displacement and does zero work on the sled in that motion.

A force can be present without doing work. The important question is not simply whether a force exists. Ask whether the object is displaced and whether that force has a component along the displacement. Use the angle between the force and displacement, not the angle between the force and some unrelated direction.

Start with the displacement direction, then compare the force to it.
Check the work of each force separately.
A force perpendicular to the displacement does no mechanical work on the object for that displacement.

4. A reliable calculation routine

First state the system, reference frame, and positive direction. Then identify the force whose work is being found, the displacement, and the angle between them. Use the force magnitude in the relationship and let the angle determine the signed result.

Substitute values with their SI units. Keep the units through the multiplication: newtons times metres gives joules. Round the final value to a sensible number of significant figures based on the given quantities.

Finally, check the result. Does its sign match the directions? Is the unit a joule? Is the size reasonable for the force and displacement? A larger force component along the displacement, or a greater displacement, should produce a greater magnitude of work.

1\,\mathrm{J}=1\,\mathrm{N\,m}

State the system and reference frame before solving.
Use the angle between force and displacement.
Report magnitude, sign, and unit; check whether the result matches the physical directions.

Worked example

A force in the direction of motion

A student pushes a storage bin with a horizontal force of

32\,\mathrm{N}

. The bin moves

4.5\,\mathrm{m}

horizontally in the direction of the push. Find the work done by the student's force.

Set the system and direction
The system is the storage bin, viewed from the ground. Choose the direction of the bin's displacement as positive. We are finding work by the student's force, not work by every force on the bin.
Identify the angle
The push and displacement point in the same direction, so the angle between them is zero degrees. The force is fully along the displacement.
$\theta=0^\circ$
Substitute and calculate
Use the work relationship. Since the cosine of zero degrees is one, the result is positive.
$W=(32\,\mathrm{N})(4.5\,\mathrm{m})\cos 0^\circ=1.44\times10^2\,\mathrm{J}$

Answer: To two significant figures, the student's force does

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

of work on the bin.

Check: The unit is

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

. The positive sign fits a force in the direction of motion. A force of a few tens of newtons acting over several metres gives work on the order of hundreds of joules, so the size is reasonable.

Worked example

A pull at an angle

A worker pulls a cart with a force of

50.\,\mathrm{N}

at

60^\circ

above the horizontal. The cart moves

3.0\,\mathrm{m}

horizontally forward. Find the work done by the pulling force.

Set the system and direction
The system is the cart, and the ground is the reference frame. Choose forward, along the cart's horizontal displacement, as positive. The force is angled above this direction.
Use the angle between the vectors
The displacement is horizontal, while the pull is $60^\circ$ above horizontal. Therefore the angle between the force and displacement is $60^\circ$ . Only the forward component of the pull contributes to work.
$\theta=60^\circ$
Substitute and calculate
Multiply the force magnitude by displacement and by the cosine of the angle. The result is positive because the force has a forward component.
$W=(50.\,\mathrm{N})(3.0\,\mathrm{m})\cos 60^\circ=75\,\mathrm{J}$

Answer: The pulling force does

75\,\mathrm{J}

of work on the cart.

Check: The unit is joules. The work is positive because part of the pull points forward. It is less than the

150\,\mathrm{J}

that a full

50.\,\mathrm{N}

horizontal pull would do over the same distance, which is reasonable because the pull is angled.

Worked example

A force opposing motion

A box slides

2.8\,\mathrm{m}

east across a floor. The friction force on the box is

12\,\mathrm{N}

west. Find the work done by friction.

Set the system and direction
The system is the box, viewed from the floor. Choose east, the direction of displacement, as positive. Friction acts west, opposite to the displacement.
Identify the angle
The friction force and displacement point in opposite directions. The angle between them is $180^\circ$ .
$\theta=180^\circ$
Substitute and calculate
The cosine of $180^\circ$ is negative one, so friction does negative work.
$W=(12\,\mathrm{N})(2.8\,\mathrm{m})\cos 180^\circ=-33.6\,\mathrm{J}$

Answer: To two significant figures, friction does

-34\,\mathrm{J}

of work on the box.

Check: The unit is joules. The negative sign agrees with friction opposing the eastward displacement. The magnitude is

12\,\mathrm{N}

times

2.8\,\mathrm{m}

, about

34\,\mathrm{J}

, which is reasonable.

Common mistakes and how to avoid them

Saying that every force on a moving object does positive work.

Correction: Compare each force direction with the displacement. A force opposing displacement does negative work; a perpendicular force does zero work.

Using the angle between a force and the horizontal even when the displacement is not horizontal.

Correction: Use the angle between the force vector and the displacement vector. These are the two directions in the work relationship.

Claiming that a force does work just because it is large or because someone is making an effort.

Correction: Mechanical work requires displacement and a force component along that displacement. If the object does not move, the force does no mechanical work on it for that interval.

Reporting a negative work value as an error.

Correction: Negative work is meaningful. It indicates that the force has a component opposite to the displacement.

Lesson summary

Mechanical work describes how a particular force acts along an object's displacement.
For a constant force and straight displacement, calculate work with $W=Fd\cos\theta$ .
Work is positive when force has a component along displacement, negative when its component opposes displacement, and zero when the force is perpendicular or there is no displacement.
Work is a scalar measured in joules.

Check your understanding

Question 1

A book moves east across a table. The table's upward normal force is perpendicular to the book's horizontal displacement. What work does the normal force do?

Positive work
Negative work
Zero work
correctIndex

Show answer and explanation

Zero work

The force is perpendicular to the displacement, so the angle is

90^\circ

and its work is zero.

Question 2

A force of

8.0\,\mathrm{N}

acts west while an object moves

5.0\,\mathrm{m}

east. What is the work done by this force?

$+40\,\mathrm{J}$
$-40\,\mathrm{J}$
$0\,\mathrm{J}$
correctIndex

Show answer and explanation

-40\,\mathrm{J}

The force is opposite to displacement, so the angle is

180^\circ

. The work is

(8.0\,\mathrm{N})(5.0\,\mathrm{m})\cos 180^\circ=-40\,\mathrm{J}

.

Question 3

A horizontal force moves a box

2.0\,\mathrm{m}

horizontally in the same direction. Which statement is correct?

The force does positive work on the box.
The force does zero work because it is horizontal.
The force does negative work because motion occurred.
correctIndex

Show answer and explanation

The force does positive work on the box.

The force points in the same direction as the displacement, so its work is positive.

Key terms

Mechanical work: A scalar measure of the effect of a force along an object's displacement.
Displacement: The change in an object's position from its starting point to its ending point, including direction.
Component: The part of a vector that points along a chosen direction.
Reference frame: The viewpoint used to describe an object's position and motion.
Scalar: A quantity with magnitude but no direction.
Vector: A quantity with both magnitude 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 D3.5. 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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D3.5 · Describe when a force does mechanical work

Ontario Grade 11 Physics

SPH3U study topic D3.5: connect force and displacement

What you will learn

1. Prerequisite bridge: force, displacement, and direction

2. The rule: compare force with displacement

3. Reading the direction before calculating

4. A reliable calculation routine

A force in the direction of motion

A pull at an angle

A force opposing motion

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