D2.2 Solve work, force, and displacement problems

Ontario Grade 11 Physics

Energy and Society

Solving Grade 11 physics problems with force and motion in a chosen direction

A force can act on an object while the object moves. Work describes the energy transferred by that force through the object's displacement. In this lesson, the system is the object being pushed or pulled. Before calculating, choose a direction and identify which force and displacement the question is asking about. Force and displacement are vectors, so each has a magnitude and direction. Work is a scalar: it has magnitude but no direction.

What you will learn

Distinguish scalar quantities from vector quantities in work problems.
Use a chosen positive direction to describe force and displacement.
Calculate work, force, or displacement when the other quantities are known.
Check units, signs, directions, significant figures, and whether an answer makes sense.

1. Review: vectors, scalars, and displacement

A scalar has magnitude only. Time and distance are examples. A vector has both magnitude and direction. Force and displacement are vectors. Displacement describes a change in position from the starting point to the ending point; it is not necessarily the total path travelled.

Choose the physical system first. Here, the system is the object whose motion is described. Then choose a positive direction, such as right or forward. A quantity in that direction is positive. A quantity in the opposite direction is negative. This sign choice lets you describe direction using numbers.

The SI unit of force is the newton, written

\mathrm{N}

. The SI unit of displacement is the metre, written

\mathrm{m}

. Work is measured in joules, written

\mathrm{J}

. One joule is equal to one newton-metre.

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

Force and displacement are vectors; work is a scalar.
State the system and positive direction before using signed values.
Use SI units: newtons, metres, and joules.

2. How force and displacement determine work

Work depends on how much of a force acts along the displacement. If the force and displacement point in the same direction, the force transfers positive work to the object. If the force points opposite to the displacement, it does negative work. If the force is at a right angle to the displacement, it does zero work in this model.

When a force is not along the displacement, use the part of the force that is parallel to the displacement. The angle in the equation is the angle between the force and displacement directions. For a force at angle

\theta

, its parallel part is

F\cos\theta

. This is a course-level trigonometry relationship for finding the side of a right triangle next to the angle.

Work can be rearranged to find an unknown force or displacement. Keep the angle information in the calculation when the force is not parallel. A negative result means the force acts against the chosen displacement; it does not mean that the amount of work has a direction.

W=Fd\cos\theta

Work is force multiplied by displacement along the force direction.
Parallel force and displacement give positive work; opposite directions give negative work.
A force perpendicular to displacement does no work in this model.

3. A reliable solution method

Write down the system, positive direction, known values, and unknown. Sketch an arrow for the displacement and an arrow for the force if the directions may be confusing. Label the angle between them when needed.

Choose the work relationship that matches the situation. If the force is parallel or opposite to displacement, use the signed direction directly. If the force is at an angle, use the angle form. Rearrange the equation before substituting when you are solving for force or displacement.

Substitute values with units. Use a sensible number of significant figures based on the given values. Check that the units reduce to joules for work, newtons for force, or metres for displacement. Finally, check that the sign and size agree with the physical situation.

F_{\parallel}=F\cos\theta

Define the direction convention before assigning positive or negative signs.
Use the force component along displacement when the directions differ.
Include units in substitutions and check the final result.

4. Interpreting the result

A large force does not always mean a large amount of work. The object must also have displacement in the force's direction. A force that is perpendicular to the displacement contributes no work, even if its magnitude is large.

Do not confuse distance with displacement. In this lesson's calculations, use the displacement from the initial position to the final position in the direction being considered. If an object moves forward while a force points backward, represent the force as opposite to the displacement.

Work is scalar, so report its value in joules and explain the sign in words when it is useful. For example, negative work means the force acted opposite to the displacement. The force itself remains a vector and should be given with a direction.

W=F_{\parallel}d

Work depends on force, displacement, and their relative direction.
A negative work value indicates opposition to the chosen displacement.
Describe force direction separately from scalar work.

Worked example

Work by a forward push

A student pushes a box with a horizontal force of

32\ \mathrm{N}

toward the right. The box moves

4.5\ \mathrm{m}

to the right. Find the work done by the student's force.

Set the direction
The system is the box. Choose right as positive. The applied force and displacement are both positive and parallel.
$F=+32\ \mathrm{N},\quad d=+4.5\ \mathrm{m},\quad \theta=0^\circ$
Choose the relationship
Work is the force component along the displacement multiplied by the displacement. Since the directions match, the force is fully along the motion.
$W=Fd\cos\theta$
Substitute and calculate
Substitute the force, displacement, and angle. The cosine of zero degrees is one, so the result is positive.
$W=(32\ \mathrm{N})(4.5\ \mathrm{m})\cos 0^\circ=144\ \mathrm{J}$

Answer: The student's force does

1.4\times10^2\ \mathrm{J}

of work on the box.

Check: The units are newton-metres, or joules. Positive work is expected because the force and displacement point right. The answer has two significant figures. A force of a few tens of newtons acting over several metres should produce work of a few hundred joules, so the magnitude is reasonable.

Worked example

Finding a resisting force

A sled moves

12\ \mathrm{m}

east. A constant force from the snow acts west and does

-180\ \mathrm{J}

of work on the sled. Find the magnitude and direction of that force.

Set the direction
The system is the sled. Choose east as positive. The displacement is positive, while the snow's force is negative because it points west.
$d=+12\ \mathrm{m},\quad W=-180\ \mathrm{J}$
Rearrange the relationship
The force is directly opposite to the displacement, so the signed relationship is work equals force multiplied by displacement. Divide by displacement to find the signed force.
$F=\frac{W}{d}$
Substitute and interpret
The negative sign gives the direction relative to east. Report the force magnitude as positive and state that its direction is west.
$F=\frac{-180\ \mathrm{J}}{+12\ \mathrm{m}}=-15\ \mathrm{N}$

Answer: The snow exerts a force of

15\ \mathrm{N}

west.

Check: Since

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

, the units are correct. The negative force is west under the chosen convention, opposite the sled's eastward displacement, which agrees with the negative work. The result has two significant figures and is reasonable: a force of

15\ \mathrm{N}

over

12\ \mathrm{m}

gives

180\ \mathrm{J}

of work in magnitude.

Worked example

Finding displacement from angled work

A person pulls a wheeled case with a force of

50\ \mathrm{N}

at

60^\circ

above the horizontal. The case moves horizontally, and the pulling force does

300\ \mathrm{J}

of work. Find the horizontal displacement.

Define the system and directions
The system is the case. Choose the horizontal direction of its motion as positive. The force is angled above that direction, so only its horizontal component contributes to work.
$F=50\ \mathrm{N},\quad \theta=60^\circ,\quad W=300\ \mathrm{J}$
Find the force component
Use the adjacent side of the force triangle to find the component along the horizontal displacement.
$F_{\parallel}=F\cos\theta=(50\ \mathrm{N})\cos 60^\circ=25\ \mathrm{N}$
Solve for displacement
Work equals the parallel force multiplied by displacement. Divide the work by the parallel force. The positive answer means the case moves in the chosen positive direction.
$d=\frac{W}{F_{\parallel}}=\frac{300\ \mathrm{J}}{25\ \mathrm{N}}=12\ \mathrm{m}$

Answer: The case moves

12\ \mathrm{m}

horizontally in the direction of the force's horizontal component.

Check: The units are joules divided by newtons, which gives metres. The vertical part of the force does not contribute to work because the displacement is horizontal. The answer is positive and reasonable: the parallel force is half of

50\ \mathrm{N}

, so it must act over a longer displacement to produce

300\ \mathrm{J}

.

Common mistakes and how to avoid them

Multiplying the full force by displacement when the force is angled.

Correction: Use only the component parallel to the displacement, or use

W=Fd\cos\theta

.

Giving a negative force magnitude.

Correction: A signed force indicates direction relative to the chosen positive direction. Report magnitude as positive and state the direction.

Assuming every force acting on a moving object does work.

Correction: A force perpendicular to displacement does zero work in this model.

Using distance travelled instead of the relevant displacement.

Correction: Identify the change in position along the direction being considered, then use that displacement.

Lesson summary

Choose the object as the system and state a positive direction.
Force and displacement are vectors; work is a scalar.
Use the component of force along displacement to calculate work.
Check units, sign, direction, significant figures, and physical reasonableness.

Check your understanding

Question 1

A force of

10\ \mathrm{N}

acts in the same direction as a displacement of

3.0\ \mathrm{m}

. How much work does the force do?

$3.0\ \mathrm{J}$
$30\ \mathrm{J}$
$-30\ \mathrm{J}$
$0\ \mathrm{J}$

Show answer and explanation

30\ \mathrm{J}

The force is parallel to displacement, so

W=Fd=(10\ \mathrm{N})(3.0\ \mathrm{m})=30\ \mathrm{J}

.

Question 2

An object moves north while a force on it points south. With north chosen as positive, what sign does the force have?

Positive, because the object is moving
Negative, because the force points opposite to positive north
Zero, because force has no direction
Positive, because force and displacement always have the same sign

Show answer and explanation

Negative, because the force points opposite to positive north

The chosen positive direction is north. A southward force points opposite to positive and is represented by a negative value.

Question 3

A force acts at right angles to an object's displacement. How much work does that force do in this model?

Positive work
Negative work
Zero work
Work equal to force times time

Show answer and explanation

Zero work

At a right angle, the force component along displacement is zero, so the force does zero work.

Key terms

System: The object or group of objects chosen for the problem.
Positive direction: The direction chosen to represent positive vector values.
Displacement: The change in position from the starting point to the ending point, including direction.
Scalar: A quantity with magnitude but no direction.
Vector: A quantity with both magnitude and direction.
Work: Energy transferred by a force acting through displacement.
Parallel component: The part of a force that points along the displacement.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation D2.2. 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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D2.2 · Solve work, force, and displacement problems

Ontario Grade 11 Physics

Solving Grade 11 physics problems with force and motion in a chosen direction

What you will learn

1. Review: vectors, scalars, and displacement

2. How force and displacement determine work

3. A reliable solution method

4. Interpreting the result

Work by a forward push

Finding a resisting force

Finding displacement from angled work

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