Work done by a variable force
Work, Energy & Power - OCR A-Level Physics
Key Idea
Work done by a varying force
The equation $W = Fx\cos\theta$ only works when $F$ is constant (or when $F$ is taken as the average force). When the force changes during the motion, the work done equals the area under the force-displacement graph.
The equation $W = Fx\cos\theta$ only works when $F$ is constant (or when $F$ is taken as the average force). When the force changes during the motion, the work done equals the area under the force-displacement graph.
$$W = \int F \, \mathrm{d}x = \text{area under } F\text{-}x \text{ graph}$$
Case 1 : Linear (Hooke's law) force
When a spring obeys Hooke's lawThe extension of a spring is proportional to the applied force, provided the limit of proportionality is not exceeded: $F = kx$., $F = kx$. The graph is a straight line through the origin, so the area under it from $0$ to extension $x$ is a triangle: $W = \tfrac{1}{2}Fx = \tfrac{1}{2}kx^{2}$. This is the elastic potential energy stored.
$$W = \tfrac{1}{2}Fx = \tfrac{1}{2}kx^{2}$$
Case 2 : Non-linear or experimental force-displacement graph
For a curved graph (e.g. a rubber band beyond its linear region, or a measured catapult force), count the squares under the curve and multiply by the area per square. Or split into strips and approximate each as a trapezium.
Diagram pending
Two force-displacement graphs side by side. Left: straight line $F = kx$ from origin, with the triangular area shaded purple and labelled $W = \tfrac{1}{2}kx^{2}$. Right: curved $F$-$x$ trace for a rubber catapult with the area under the curve shaded and labelled "work done = area under graph"; small grid squares visible to show the square-counting method.
Will be replaced with a GeoGebra SVG in stream 2.
- The area-under-graph method works for any force that varies with displacement, including springs, rubber bands, gravity at large distances, and electrostatic forces.
- If the force opposes motion (e.g. friction acting backwards), the area is counted as negative work.
- For a spring, the work done on the spring (storing elastic PE) equals the work done by the spring on release. Energy is recoverable.
- For a rubber band, the loading and unloading curves differ. The area between them is energy lost to thermal stores (hysteresis).
- Quick check: if $F$ is constant, the graph is a horizontal line and the area is a rectangle $Fx$, recovering $W = Fx$.
Common Mistake
HIGH
Wrong: Using $W = Fx$ with the maximum force on a spring graph. For a spring stretched to extension $x$ with maximum force $F$, this gives double the actual work done.
Right: The force varies linearly from $0$ to $F$, so use the average: $W = \tfrac{1}{2}Fx$. Equivalently, the area under the $F$-$x$ graph is a triangle, not a rectangle.
Right: The force varies linearly from $0$ to $F$, so use the average: $W = \tfrac{1}{2}Fx$. Equivalently, the area under the $F$-$x$ graph is a triangle, not a rectangle.
Worked Example [3 marks]
A student stretches a catapult elastic band that obeys Hooke's law with spring constant $k = 250 \text{ N m}^{-1}$ by $0.20 \text{ m}$. They release a $10 \text{ g}$ pellet. Assuming all the stored elastic energy transfers to kinetic energy of the pellet, calculate the launch speed.
Show Solution
1
Elastic PE stored
$W = \tfrac{1}{2}kx^{2} = \tfrac{1}{2} \times 250 \times 0.20^{2} = 5.0 \text{ J}$.
[1]2
All elastic PE becomes KE
$\tfrac{1}{2}mv^{2} = 5.0$, so $v^{2} = \frac{2 \times 5.0}{0.010} = 1000 \text{ m}^{2} \text{ s}^{-2}$.
[1]3
$v = \sqrt{1000} \approx 32 \text{ m s}^{-1}$.
[1]Answer
$32 \text{ m s}^{-1}$
Examiner Tips and Tricks
- When given an experimental $F$-$x$ graph, mark the area you are counting clearly. Examiners often credit the method (area under graph) even if the arithmetic is slightly off.
- If the question mentions a "loading" and an "unloading" curve for an elastic band, expect a follow-up about the energy transferred to thermal stores. The difference of areas is the dissipated energy.