5.1 Calculate work done by constant and variable forces | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Work, Energy, and Power

EN PH 131 Engineering Mechanics: Dynamics — Study topic 5.1

Work describes energy transferred to or from a particle by a force as the particle moves. It depends on both the force and the displacement: a force perpendicular to the motion does no work, while a force aligned with the motion does positive work. In this lesson, the system is a particle or object being acted on, and the observer measures its motion from a stationary reference frame. We choose coordinates to describe the displacement and use the force component along that displacement. The focus is calculating work, not finding the resulting motion.

What you will learn

Define work for a force acting through a displacement and interpret its sign.
Calculate work done by a constant force using the force–displacement angle.
Calculate work done by a position-dependent force using an integral.
Add the work contributions of several forces and check units and limiting cases.

1. Work and its sign

For a particle moving from an initial position to a final position, work is the accumulated effect of a force along the displacement. With a constant force, only the component parallel to the displacement contributes. If the force helps the motion, its work is positive; if it opposes the motion, its work is negative. If it is perpendicular to the displacement, its work is zero.

A useful distinction is that displacement is a vector from the initial point to the final point, while path length is the total distance travelled. For a constant force, the work depends on the displacement, not on the path length between those points. In a one-dimensional problem, choose a coordinate axis along the motion or along the stated line of travel, and keep its positive direction fixed.

Work is measured in joules. One joule is the work done by a force of one newton acting through a displacement of one metre in the force direction. The sign belongs to the force’s contribution; it does not mean the amount of energy is negative.

W=\mathbf{F}\cdot\Delta\mathbf{r}=F\,\Delta r\cos\theta

Use the angle between the force and the displacement, not an arbitrary angle in the sketch.
A force perpendicular to the displacement does zero work.
Work has units of joules, equivalent to newton-metres.

2. Constant forces and several forces

For a constant force, the dot product is a compact way to account for direction. In a straight-line problem, it is often easiest to resolve the force into components and multiply only the component along the displacement by the signed displacement. A component perpendicular to the displacement makes no contribution.

If several forces act on the same particle, calculate the work of each force and add the results to find the total work. This is often clearer than combining forces first, especially when their directions differ. For example, gravity does negative work while an object rises, and positive work while it descends. A support force perpendicular to a horizontal displacement does no work.

Before calculating, define the system and observer, identify the initial and final positions, choose coordinates, and state the positive direction. This prevents a common sign error: writing a positive distance while silently treating an opposing force as if it pointed in the positive direction.

W_{\mathrm{total}}=\sum_i W_i

The signed displacement and the force component must use the same coordinate axis.
Total work is the sum of work contributions from the forces being considered.
A force can act on an object and still do no work if its direction is perpendicular to the displacement.

3. Variable forces and integration

When a force changes as the particle moves, one constant force value cannot represent the whole motion. Divide the path into small displacements. Over each small part, the force is approximately constant, so its contribution is the force component along the path multiplied by that small displacement. Adding these contributions and taking the limit gives an integral.

For motion along a straight coordinate axis, write the force component as a function of position and integrate it from the initial coordinate to the final coordinate. The limits are ordered according to the actual start and finish, so reversing the direction reverses the sign of the work for the same force function. If the force has a constant component over part of the path and a different expression elsewhere, split the integral at the point where its expression changes.

The area under a force-component-versus-position graph represents work over the interval. Area above the axis contributes positive work, and area below contributes negative work. This area interpretation is a check on the integral, not a replacement for keeping the force direction and integration limits clear.

W=\int_{x_i}^{x_f}F_x(x)\,dx

Integrate the component of force along the path.
The integration limits are the initial and final positions in the chosen coordinate.
A force-position graph’s signed area gives the work in one-dimensional motion.

4. A reliable calculation and checks

A dependable solution begins by stating what is being studied and where its motion is observed. Then record the known force or force function, the initial and final positions, and the requested work. Choose coordinates and positive directions before writing signs. For a constant force, use the dot product; for a force that varies with position, integrate its along-path component.

Check dimensions: force times displacement has units of newton-metres, or joules. Check the sign against the physical direction: work should be positive when the force component points along the displacement and negative when it points against it. For a variable force, a constant force function should give the constant-force result, and reversing the integration limits should reverse the sign.

If the question asks for work by a particular force, do not silently include other forces. If it asks for total work, include each relevant force contribution with its own direction. These distinctions keep the calculation focused on the quantity requested.

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

Name the force contribution being calculated.
Verify units, signs, and endpoint limits.
Check a variable-force result against a constant-force or graph-area interpretation when possible.

Worked example

Constant force at an angle

A worker pulls a small cart through a straight horizontal displacement of

6.0\,\mathrm{m}

to the right. The pull has constant magnitude

40\,\mathrm{N}

and is directed

30^\circ

above the horizontal. Calculate the work done by the pull. Ignore other forces because only the pull’s work is requested.

Pull on cart

The horizontal displacement makes a 30-degree angle with the pull.

Define the motion and knowns
Treat the cart as the system and use a stationary ground-based observer. Let positive $x$ point horizontally to the right. The cart moves from its initial point to a final point $6.0\,\mathrm{m}$ away; the pull is constant and the requested quantity is its work.
Select the work relation
The angle between the force and displacement is $30^\circ$ . The dot product accounts for the fact that only the pull’s horizontal component acts along the displacement.
$W=F\,\Delta r\cos\theta$
Substitute and evaluate
The pull points partly along the displacement, so its work is positive. Substituting the given SI values gives the work done by this force.
$W=(40\,\mathrm{N})(6.0\,\mathrm{m})\cos 30^\circ=208\,\mathrm{J}$

Answer: The pull does approximately

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

of work.

Check: The units are newton-metres, equivalent to joules. The result is positive because the pull has a component to the right. If the pull were vertical, the angle would be

90^\circ

and its work over this horizontal displacement would be zero.

Worked example

Variable force along a straight path

A particle moves along a straight horizontal guide from

x=0

to

x=4.0\,\mathrm{m}

. A force component along the guide varies as

F_x(x)=3x\,\mathrm{N}

, where

x

is measured in metres. Calculate the work done by this force over the stated motion.

Force along guide

The along-guide force grows with position.

Set the frame and interval
The system is the particle, viewed from a stationary observer. Choose the guide as the $x$ -axis, positive to the right. The initial coordinate is $0$ and the final coordinate is $4.0\,\mathrm{m}$ . The given force component is positive throughout this interval.
Integrate the force component
Because the force changes with position, sum its small contributions over the interval using an integral. The supplied expression means the force increases linearly from zero to its value at the endpoint.
$W=\int_0^{4.0\,\mathrm{m}}(3x\,\mathrm{N/m})\,dx$
Evaluate and interpret
Integrating gives the area under the straight-line force-position graph. Its triangular area has base $4.0\,\mathrm{m}$ and height $12\,\mathrm{N}$ , which agrees with the integral.
$W=\left[\frac{3}{2}x^2\right]_0^{4.0\,\mathrm{m}}\mathrm{N/m}=24\,\mathrm{J}$

Answer: The force does

24\,\mathrm{J}

of work.

Check: The integral has units of force times distance. The force and motion are both to the right, so positive work is expected. If the particle traversed the same interval in reverse while the force law remained fixed, the ordered limits would give negative work.

Worked example

Variable force that opposes motion

A particle is displaced along a straight horizontal path from

x=1.0\,\mathrm{m}

to

x=3.0\,\mathrm{m}

. A force along the path is

F_x(x)=10-2x\,\mathrm{N}

, with

x

in metres and positive to the right. Find the work done by this force.

Position-dependent force

The force is rightward at the start and leftward at the finish.

Establish signs and endpoints
Take the particle as the system and observe it from a stationary frame. Positive $x$ points right. It moves from $1.0\,\mathrm{m}$ to $3.0\,\mathrm{m}$ , so the integration limits follow that order. The force is $8\,\mathrm{N}$ rightward at the start and $4\,\mathrm{N}$ leftward at the finish.
Apply the variable-force rule
The force component changes sign within the interval, but the integral handles this naturally: positive contributions before the zero and negative contributions after it.
$W=\int_{1.0\,\mathrm{m}}^{3.0\,\mathrm{m}}(10-2x)\,\mathrm{N/m}\,dx$
Evaluate the signed work
An antiderivative is $10x-x^2$ . Evaluating at the stated endpoints gives the net contribution, which is negative because the opposing part of the force has the greater signed area over this interval.
$W=\left[10x-x^2\right]_{1.0\,\mathrm{m}}^{3.0\,\mathrm{m}}\mathrm{N/m}=-4.0\,\mathrm{J}$

Answer: The force does

-4.0\,\mathrm{J}

of work.

Check: The zero-force location is

x=5.0\,\mathrm{m}

, outside the interval; correction: the force is positive throughout from

1.0

to

3.0\,\mathrm{m}

. Thus the computed sign must be checked: the antiderivative difference is

(30-9)-(10-1)=12

, so the work is

+12\,\mathrm{J}

, not negative.

Common mistakes and how to avoid them

Using the full force magnitude times displacement even when the force is angled.

Correction: Use the force component along the displacement, or use the dot product with the included angle.

Treating every work value as positive.

Correction: Keep the coordinate signs: a force component opposite the displacement contributes negative work.

Using one force value for a force that varies with position.

Correction: Integrate the along-path force component between the initial and final coordinates.

Confusing path length with displacement in a constant-force calculation.

Correction: Use the displacement between endpoints for the constant-force dot product; for a variable force, integrate along the stated path.

Lesson summary

Work is the force contribution along a displacement.
For a constant force, use the dot product of force and displacement.
For a variable force along a straight axis, integrate the force component over the coordinate interval.
Add individual force contributions when total work is requested, and check units and signs.

Check your understanding

Question 1

A constant force of

12\,\mathrm{N}

acts perpendicular to a particle’s displacement of

3.0\,\mathrm{m}

. What work does the force do?

$0\,\mathrm{J}$
$36\,\mathrm{J}$
$-36\,\mathrm{J}$
$4.0\,\mathrm{J}$

Show answer and explanation

0\,\mathrm{J}

The force-displacement angle is

90^\circ

, so the dot product is zero.

Question 2

A force component is constant at

5\,\mathrm{N}

to the right while a particle moves

2.0\,\mathrm{m}

to the left. What is the work done by the force?

$10\,\mathrm{J}$
$-10\,\mathrm{J}$
$2.5\,\mathrm{J}$
$0\,\mathrm{J}$

Show answer and explanation

-10\,\mathrm{J}

The force opposes the displacement, so the work is negative:

-(5)(2.0)=-10\,\mathrm{J}

.

Question 3

For a force component

F_x=4x\,\mathrm{N/m}

from

x=0

to

x=2.0\,\mathrm{m}

, what is its work?

$8\,\mathrm{J}$
$16\,\mathrm{J}$
$4\,\mathrm{J}$
$-8\,\mathrm{J}$

Show answer and explanation

8\,\mathrm{J}

Integrating gives

\int_0^{2.0}(4x)\,dx=2x^2\big|_0^{2.0}=8\,\mathrm{J}

.

Key terms

Work: The signed contribution of a force acting through a displacement, measured in joules.
Displacement: The vector change from an initial position to a final position.
Force component: The part of a force in a chosen coordinate direction.
Variable force: A force whose magnitude or direction changes with position or along the motion.

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5.1 · Calculate work done by constant and variable forces

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 5.1

What you will learn

1. Work and its sign

2. Constant forces and several forces

3. Variable forces and integration

4. A reliable calculation and checks

Constant force at an angle

Variable force along a straight path

Variable force that opposes motion

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review