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Work and Variable Forces · Tutorial 452 of 998

Work from Force-Position Graphs

Interpret the signed area between a force-position graph and the position axis as work, including when the graph crosses the axis or motion reverses.

Intermediate 9 min read

What You'll Learn

  • Interpret positive and negative regions of a force-position graph as positive and negative work.
  • Calculate work from triangular, rectangular, and trapezoidal regions.
  • Find and use the axis crossing of a straight graph segment.
  • Add signed areas across multiple linear segments.
  • Account for the sign change when motion proceeds toward decreasing position.
  • Check a piecewise-linear work calculation using endpoint forces.

Reading Work from a Graph

A force-position graph can show how much work a force does without requiring you to start with a formula for the force. The key is to read the area between the graph and the position axis with its sign: regions above the axis contribute positive work, and regions below it contribute negative work.

This is a signed-area calculation, not a calculation of the total geometric area. If a graph crosses the axis, its regions on opposite sides must be counted with opposite signs.

Definition: For a force \(F(x)\) along the position axis, the work done as an object moves from position \(x_i\) to position \(x_f\) is represented on a force-position graph by the signed area between the graph and the axis from \(x_i\) to \(x_f\). Area above the axis is positive; area below it is negative. Work is measured in joules, where \(1\text{ J}=1\text{ N}\cdot\text{m}\).

This interpretation follows the sign convention for work: a force in the direction of motion does positive work, while a force opposing the motion does negative work. Here, the graph gives the signed force component along the position axis. The interpretation applies when that component is plotted against position and the motion is along the axis.

Areas of Basic Shapes

For a straight graph segment, the signed area between two positions can be calculated as a trapezoid. The endpoint force values determine its two parallel sides. If one endpoint force is zero, the shape is a triangle; if both forces are equal, it is a rectangle.

$$ W=\frac{x_b-x_a}{2}\bigl(F(x_a)+F(x_b)\bigr) $$

This formula uses the signed position difference \(x_b-x_a\). For motion toward increasing position, the width is positive. For motion toward decreasing position, it is negative. The formula assumes the graph is a straight line between the two endpoints; for a piecewise-linear graph, apply it separately on each straight segment and add the results.

Useful area formulas: A triangle has area \(\frac12 bh\), a rectangle has area \(bh\), and a trapezoid with parallel sides \(h_1\) and \(h_2\) has area \(\frac12 b(h_1+h_2)\). For work from a graph, assign a positive sign to area above the position axis and a negative sign to area below it.

Why the Trapezoid Formula Works

On a straight graph segment, the force changes linearly from one endpoint value to the other. The signed area under that segment is the area of a trapezoid: its width is the change in position, and its parallel sides are the endpoint force values. The trapezoid area is the width times the average of those two sides.

$$ \text{signed area} =(x_b-x_a)\frac{F(x_a)+F(x_b)}{2} $$

If the line crosses the position axis, the same formula still gives the net signed area. The part below the axis is subtracted from the part above it. Thus the formula works whether both endpoint forces have the same sign or have opposite signs. It also works for reversed motion because exchanging the endpoints changes the sign of \(x_b-x_a\).

For a graph made of several straight segments, divide the interval at every vertex. Signed areas add across adjacent intervals, so the total work is the sum of the segment contributions. This is why it is important to include every change of slope and every crossing when partitioning a graph.

Worked Example: A Triangle, a Rectangle, and a Triangle

Worked Example: A Triangle, a Rectangle, and a Triangle

A force-position graph runs from \((0\text{ m},0\text{ N})\) to \((3\text{ m},6\text{ N})\), remains at \(6\text{ N}\) until \(x=5\text{ m}\), and then falls linearly to \((8\text{ m},0\text{ N})\). Find the work as the object moves from \(0\text{ m}\) to \(8\text{ m}\).

The graph consists of a triangle from \(0\) to \(3\text{ m}\), a rectangle from \(3\) to \(5\text{ m}\), and a triangle from \(5\) to \(8\text{ m}\). All three regions are above the axis, so each contributes positive work.

$$ W_1=\frac12(3\text{ m})(6\text{ N})=9\text{ J} $$

The rectangle has width \(5-3=2\text{ m}\) and height \(6\text{ N}\):

$$ W_2=(2\text{ m})(6\text{ N})=12\text{ J} $$

The final triangle has width \(8-5=3\text{ m}\) and height \(6\text{ N}\):

$$ W_3=\frac12(3\text{ m})(6\text{ N})=9\text{ J} $$

Adding the contributions gives

$$ W=W_1+W_2+W_3 =9\text{ J}+12\text{ J}+9\text{ J} =30\text{ J}. $$

Check: The straight-segment trapezoid formula gives \(3(0+6)/2=9\text{ J}\) on the first segment, \(2(6+6)/2=12\text{ J}\) on the middle segment, and \(3(6+0)/2=9\text{ J}\) on the final segment. These add to \(30\text{ J}\), matching the geometric calculation.

Worked Example: A Piecewise-Linear Graph with Crossings

Worked Example: A Piecewise-Linear Graph with Crossings

A graph connects the points \((0\text{ m},4\text{ N})\), \((3\text{ m},-2\text{ N})\), and \((5\text{ m},2\text{ N})\) with straight segments. Find the work from \(x=0\text{ m}\) to \(x=5\text{ m}\).

First find where each segment crosses the position axis. On the segment from \(x=0\) to \(x=3\), the force drops by \(6\text{ N}\) over \(3\text{ m}\), a rate of \(-2\text{ N/m}\). Starting at \(4\text{ N}\), it reaches zero after \(2\text{ m}\), so the crossing is at \(x=2\text{ m}\).

On the segment from \(x=3\) to \(x=5\), the force rises from \(-2\text{ N}\) to \(2\text{ N}\). It reaches zero halfway along, at \(x=4\text{ m}\). The graph therefore forms four triangles: a positive triangle from \(0\) to \(2\), a negative one from \(2\) to \(3\), a negative one from \(3\) to \(4\), and a positive one from \(4\) to \(5\).

$$ W_1=\frac12(2\text{ m})(4\text{ N})=4\text{ J} $$
$$ W_2=-\frac12(1\text{ m})(2\text{ N})=-1\text{ J} $$
$$ W_3=-\frac12(1\text{ m})(2\text{ N})=-1\text{ J} $$
$$ W_4=\frac12(1\text{ m})(2\text{ N})=1\text{ J} $$

Thus the net work is

$$ W=4\text{ J}-1\text{ J}-1\text{ J}+1\text{ J}=3\text{ J}. $$

Check: Apply the trapezoid formula directly to the two original segments. The first contributes \(3(4+(-2))/2=3\text{ J}\). The second contributes \(2((-2)+2)/2=0\text{ J}\). Their sum is \(3\text{ J}\), as found by splitting at the crossings.

Worked Example: Work When Position Decreases

Worked Example: Work When Position Decreases

On the interval from \(x=1\text{ m}\) to \(x=4\text{ m}\), a straight force-position graph rises from \(2\text{ N}\) to \(8\text{ N}\). An object moves from \(x=4\text{ m}\) back to \(x=1\text{ m}\). Find the work done by the force.

The graph lies above the axis. The area between \(1\text{ m}\) and \(4\text{ m}\) has width \(3\text{ m}\) and parallel sides \(2\text{ N}\) and \(8\text{ N}\), so its magnitude is

$$ \frac12(3\text{ m})(2\text{ N}+8\text{ N})=15\text{ J}. $$

But the object moves from the larger position to the smaller one. The signed position change is \(1-4=-3\text{ m}\), so the work is negative:

$$ W=\frac{1-4}{2}(8+2)\text{ N}\cdot\text{m} =-15\text{ J}. $$

Check: If the same force acted while the object moved from \(1\text{ m}\) to \(4\text{ m}\), the signed area would be \(+15\text{ J}\). Reversing the direction of motion reverses the sign, giving \(-15\text{ J}\) for the stated motion.

A Common Mistake: Adding Areas Without Their Signs

In the crossing example, the first segment has endpoint forces \(4\text{ N}\) and \(-2\text{ N}\). A common error is to replace the negative force with its magnitude and calculate \(3(4+2)/2=9\text{ J}\). This treats the area below the axis as positive. It is not the net work.

The correct calculation keeps the force sign:

$$ \frac{3}{2}\bigl(4+(-2)\bigr)=3\text{ J}. $$

Alternatively, split the segment at its crossing: the positive triangle contributes \(4\text{ J}\), and the negative triangle contributes \(-1\text{ J}\), for a net \(3\text{ J}\). Both methods give the same signed area. Using absolute values is appropriate only if the question asks for total geometric area, not net work.

A Reliable Graph-Reading Process

Before calculating, mark the initial and final positions, all graph vertices between them, and every crossing of the position axis. Then assign each region its sign. The following sequence helps prevent missed segments and sign errors.

1
Set the interval and direction.
Identify the starting and ending positions. Their order determines the sign associated with the direction of motion.
2
Partition the graph.
Divide at every vertex and every axis crossing so that each piece has a simple shape and a consistent sign.
3
Calculate signed areas.
Use triangle, rectangle, or trapezoid areas. Keep force values below the axis negative.
4
Add and check.
Sum the signed contributions and compare with the endpoint trapezoid formula on each straight segment.

Check Your Understanding

Use signed areas throughout. Include units in each final answer.

  1. A constant force of \(5\text{ N}\) acts from \(x=1\text{ m}\) to \(x=4\text{ m}\). What is the work for motion toward increasing position?
  2. A straight graph rises from \(0\text{ N}\) to \(6\text{ N}\) over \(2\text{ m}\). What work does it represent over that interval?
  3. A straight segment runs from \((0\text{ m},4\text{ N})\) to \((4\text{ m},-4\text{ N})\). Where does it cross the axis, and what is the net work from \(0\text{ m}\) to \(4\text{ m}\)?
  4. A straight graph has endpoint forces \(2\text{ N}\) and \(6\text{ N}\) between positions \(1\text{ m}\) and \(3\text{ m}\). Find the work for motion from \(3\text{ m}\) to \(1\text{ m}\).
  5. When a graph has regions both above and below the position axis, why is adding their unsigned areas generally not the work?

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