C2.2 Solve one- and two-dimensional work-energy problems

Ontario Grade 12 Physics

Energy and Momentum

Use force, displacement, and the change in kinetic energy to solve motion problems.

Work–energy problems connect forces acting over a displacement to changes in motion. The work done by one force depends on both that force and the object's displacement. The net work done by all forces changes the object's kinetic energy. Before calculating, identify the system—the object or objects being studied—and the reference frame, the viewpoint used to describe motion. In this lesson, use the ground as the reference frame unless stated otherwise. Choose a positive direction that matches the displacement when possible. Force and displacement are vectors: each has magnitude and direction. Work and kinetic energy are scalars: each has magnitude but no direction.

What you will learn

Identify the system, reference frame, positive direction, and displacement in a work–energy problem.
Calculate work from a force and displacement, including when they point in different directions.
Use net work to find a change in kinetic energy or an object's final speed.
Check the sign, units, direction, and reasonableness of a result.

1. From forces and displacement to work

A displacement is a change in position. It is a vector measured in metres. A force is also a vector, measured in newtons. Work is the energy transferred by a force as an object moves through a displacement. Work is measured in joules, where one joule is one newton-metre.

Only the part of a force along the displacement contributes to its work. If the force points along the displacement, the work is positive. If it points against the displacement, the work is negative. If it is perpendicular to the displacement, it does no work. The angle in the work equation is the angle between the force and displacement vectors.

In two dimensions, resolve vectors into horizontal and vertical components. For example, a force at an angle can be written using its horizontal component and vertical component. Multiply matching components of force and displacement, then add. This is a component method for finding work and avoids treating a vector as if it were a scalar.

W=Fd\cos\theta=F_x\Delta x+F_y\Delta y

Positive work transfers energy to the object's motion; negative work removes energy from it.
A force perpendicular to the displacement does zero work.
Use the same coordinate directions for all vector components.

2. Net work and kinetic energy

Kinetic energy is the energy an object has because it is moving. It is a scalar and is measured in joules. Its value depends on the object's mass and speed, not on the direction of travel. Speed is the magnitude of velocity.

The work–energy relationship states that the net work done on a system equals its change in kinetic energy. Net work means the sum of the work done by every force on the system. Include positive and negative contributions with their signs. The relationship applies whether the motion is in one dimension or two.

Choose the system carefully. If the system is a single object, include the work done on it by forces such as an applied force, friction, or gravity. For gravity near Earth's surface, the weight is the gravitational force and points downward. A normal force is the support force from a surface; when it is perpendicular to the object's displacement, it does no work.

This method can find a final speed without first finding the time taken or the acceleration. If a final speed is requested, solve the work–energy relationship for it and use the non-negative speed. The direction of motion must come from the problem's displacement or other stated information, not from kinetic energy alone.

W_{\mathrm{net}}=\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2

Add the work of all forces to obtain net work.
Kinetic energy is never negative because it depends on speed squared.
A result for speed gives magnitude; state direction separately when the information allows it.

3. A reliable problem-solving method

First name the system and reference frame. Then choose and state positive directions. Record the known mass, forces, displacement, and initial speed, and identify the unknown. A force diagram can help list the forces, but work depends on how each force is oriented relative to the displacement.

Next, describe the displacement with components if it is not along one axis. Find the work of each force using the component form or the angle form. Keep the sign of each contribution. Add the contributions to find net work, then use the change in kinetic energy relationship.

Keep SI units in substitutions: mass in kilograms, speed in metres per second, force in newtons, displacement in metres, and work or energy in joules. Round the final result to a sensible number of significant figures based on the given values. Finally, check that the work has units of joules, the answer has the requested units, and the signs and size make physical sense.

1\ \mathrm{J}=1\ \mathrm{N\,m}=1\ \mathrm{kg\,m^2/s^2}

A negative net work means kinetic energy decreases, not that kinetic energy itself is negative.
Use the component of force parallel to displacement.
Check whether the object's speed should increase or decrease based on the net work.

Worked example

One-dimensional motion with friction

A

3.0\ \mathrm{kg}

cart starts from rest and moves

5.0\ \mathrm{m}

to the right on a level track. A

18\ \mathrm{N}

applied force acts right, while a

4.0\ \mathrm{N}

friction force acts left. Find the cart's final speed.

Set the system and direction
Take the cart as the system and the ground as the reference frame. Choose right as positive. The displacement is right, the initial speed is zero, and the unknown is the final speed.
Find each force's work
The applied force points with the displacement, so its work is positive. Friction points against the displacement, so its work is negative. Both forces act along the one-dimensional path.
$W_{\mathrm{applied}}=(18\ \mathrm{N})(5.0\ \mathrm{m})=90\ \mathrm{J},\quad W_{\mathrm{friction}}=-(4.0\ \mathrm{N})(5.0\ \mathrm{m})=-20\ \mathrm{J}$
Use net work
The net work is the sum of the two contributions. Apply the work–energy relationship and substitute the initial speed of zero.
$W_{\mathrm{net}}=90\ \mathrm{J}-20\ \mathrm{J}=70\ \mathrm{J}=\frac{1}{2}(3.0\ \mathrm{kg})v_f^2-0$
Solve and check
Solving for speed gives a positive magnitude. The units reduce to metres per second. Positive net work means the cart gains kinetic energy, so a nonzero final speed is reasonable.
$v_f=\sqrt{\frac{2(70\ \mathrm{J})}{3.0\ \mathrm{kg}}}=6.8\ \mathrm{m/s}$

Answer: The cart's final speed is

6.8\ \mathrm{m/s}

to the right.

Check: The net work is positive, so the cart speeds up from rest. Also,

\mathrm{J/kg}

has units of

\mathrm{m^2/s^2}

, as required before taking the square root.

Worked example

Two-dimensional force and displacement

A

2.0\ \mathrm{kg}

object has an initial speed of

3.0\ \mathrm{m/s}

. It moves through a displacement of

3.0\ \mathrm{m}

east and

4.0\ \mathrm{m}

north while a constant

6.0\ \mathrm{N}

east force and a constant

8.0\ \mathrm{N}

north force act on it. Find its final speed, assuming these are the only forces doing work.

Set axes and known values
Use the object as the system and the ground as the reference frame. Choose east as positive $x$ and north as positive $y$ . The force and displacement components have the same signs on both axes.
$\Delta x=3.0\ \mathrm{m},\quad \Delta y=4.0\ \mathrm{m},\quad F_x=6.0\ \mathrm{N},\quad F_y=8.0\ \mathrm{N}$
Calculate the work
Add the products of matching components. Both forces act in the direction of their corresponding displacement components, so both work contributions are positive.
$W_{\mathrm{net}}=(6.0\ \mathrm{N})(3.0\ \mathrm{m})+(8.0\ \mathrm{N})(4.0\ \mathrm{m})=50\ \mathrm{J}$
Apply work–energy
The initial kinetic energy is included because the object is already moving. Substitute the net work and solve for the final speed.
$50\ \mathrm{J}=\frac{1}{2}(2.0\ \mathrm{kg})v_f^2-\frac{1}{2}(2.0\ \mathrm{kg})(3.0\ \mathrm{m/s})^2$
Find the speed and assess it
Rearranging gives the final speed. The answer is greater than the initial speed, which agrees with the positive net work. Kinetic energy alone does not determine the final direction.
$v_f=\sqrt{(3.0\ \mathrm{m/s})^2+\frac{2(50\ \mathrm{J})}{2.0\ \mathrm{kg}}}=7.7\ \mathrm{m/s}$

Answer: The final speed is

7.7\ \mathrm{m/s}

. Its direction is not determined by this energy calculation alone.

Check: The displacement and force components are in metres and newtons, so each product is in joules. The speed increased from

3.0\ \mathrm{m/s}

, consistent with positive net work.

Worked example

Two-dimensional motion with opposing work

A

2.0\ \mathrm{kg}

object moves through a displacement of

3.0\ \mathrm{m}

east and

4.0\ \mathrm{m}

north. Its initial speed is

1.0\ \mathrm{m/s}

. An applied force has components

12\ \mathrm{N}

east and

5.0\ \mathrm{N}

north. A second force of

2.0\ \mathrm{N}

acts west. Find the final speed.

Choose the system and axes
Take the object as the system and use the ground as the reference frame. East is positive $x$ and north is positive $y$ . The westward force has a negative $x$ component.
$\Delta x=3.0\ \mathrm{m},\quad \Delta y=4.0\ \mathrm{m},\quad F_{x,\mathrm{west}}=-2.0\ \mathrm{N}$
Find the applied-force work
The applied force has east and north components. Multiply each by displacement along the same axis and add the results.
$W_{\mathrm{applied}}=(12\ \mathrm{N})(3.0\ \mathrm{m})+(5.0\ \mathrm{N})(4.0\ \mathrm{m})=56\ \mathrm{J}$
Include opposing work
The westward force is opposite the eastward displacement, so its work is negative. There is no northward component for this force.
$W_{\mathrm{west}}=(-2.0\ \mathrm{N})(3.0\ \mathrm{m})=-6.0\ \mathrm{J},\quad W_{\mathrm{net}}=50\ \mathrm{J}$
Calculate final speed
Use the initial kinetic energy and net work. The positive net work raises kinetic energy, and the resulting speed is physically possible.
$v_f=\sqrt{(1.0\ \mathrm{m/s})^2+\frac{2(50\ \mathrm{J})}{2.0\ \mathrm{kg}}}=7.1\ \mathrm{m/s}$

Answer: The object's final speed is

7.1\ \mathrm{m/s}

.

Check: The opposing force reduces the applied-force work from

56\ \mathrm{J}

to a net

50\ \mathrm{J}

. The final speed is therefore lower than it would be if the westward force were absent.

Common mistakes and how to avoid them

Using the full force magnitude when the force is angled to the displacement.

Correction: Use the force component along the displacement, or use the angle between the force and displacement in the work equation.

Adding the magnitudes of work contributions without signs.

Correction: Give positive signs to work along the displacement and negative signs to work against it before adding.

Treating a negative net work value as negative kinetic energy.

Correction: Negative net work means kinetic energy decreases. Kinetic energy itself remains zero or positive.

Assuming work–energy gives the direction of the final velocity.

Correction: Work–energy gives a speed. State the direction only when the motion information or another relationship establishes it.

Lesson summary

Define the system, reference frame, positive axes, and displacement before solving.
Calculate work using the force component along the displacement.
Sum the signed work of all forces, then set net work equal to the change in kinetic energy.
Check units, signs, significant figures, and whether the result matches the expected change in speed.

Check your understanding

Question 1

A

10\ \mathrm{N}

force acts perpendicular to a

2.0\ \mathrm{m}

displacement. What work does this force do?

$0\ \mathrm{J}$
$20\ \mathrm{J}$
$-20\ \mathrm{J}$
correctIndex 0 5

Show answer and explanation

0\ \mathrm{J}

A perpendicular force has no component along the displacement, so it does zero work.

Question 2

An object's net work is

-12\ \mathrm{J}

. What does this tell you about its kinetic energy?

Its kinetic energy decreases by $12\ \mathrm{J}$ .
Its kinetic energy becomes $-12\ \mathrm{J}$ .
Its speed must become zero.
correctIndex 0 5

Show answer and explanation

Its kinetic energy decreases by

12\ \mathrm{J}

.

Net work equals the change in kinetic energy. Negative net work means a decrease of

12\ \mathrm{J}

, but it does not by itself mean the object stops.

Question 3

A

2.0\ \mathrm{kg}

object starts from rest. A net work of

16\ \mathrm{J}

is done on it. What is its final speed?

$4.0\ \mathrm{m/s}$
$8.0\ \mathrm{m/s}$
$2.0\ \mathrm{m/s}$
correctIndex 0 5

Show answer and explanation

4.0\ \mathrm{m/s}

Using net work equal to the change in kinetic energy gives

16\ \mathrm{J}=\frac{1}{2}(2.0\ \mathrm{kg})v_f^2

, so the final speed is

4.0\ \mathrm{m/s}

.

Key terms

System: The object or objects selected for study.
Reference frame: The viewpoint used to describe an object's position and motion.
Displacement: The vector change in an object's position.
Work: Energy transferred by a force acting through a displacement.
Net work: The signed sum of the work done by all forces on the system.
Kinetic energy: The energy an object has because it is moving.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation C2.2. 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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C2.2 · Solve one- and two-dimensional work-energy problems

Ontario Grade 12 Physics

Use force, displacement, and the change in kinetic energy to solve motion problems.

What you will learn

1. From forces and displacement to work

2. Net work and kinetic energy

3. A reliable problem-solving method

One-dimensional motion with friction

Two-dimensional force and displacement

Two-dimensional motion with opposing work

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH4U

About this lesson and its review