Work done by a variable force

How do you find the work done by a variable force?

In plain words: when the force changes as the body moves, you cannot simply multiply one force by the distance. Instead, cut the motion into many tiny steps, treat the force as constant over each step, add up the small pieces of work, and let the steps shrink to zero. The result is the area under the force–displacement graph, which in calculus is a definite integral.

For a force F(x) acting along the x-axis, as a body moves from x_i to x_f:

A constant force is the special case where the graph is a horizontal line and the area is just a rectangle, F × (x_f − x_i).

Work as the area under the graph

Over a very small displacement Δx, the force hardly changes, so the work done is close to that of a constant force:

If the displacement ∆x is small, we can take the force F(x) as approximately constant and the work done is then ∆W =F(x) ∆x

NCERT Physics Part I (Class 11), Section 5.5, page 75

On a graph of F against x, this small piece of work is the area of a thin rectangle of height F(x) and width Δx. Adding all such strips from x_i to x_f gives approximately the total work:

As the strips are made thinner (Δx → 0), their number grows without limit, but the sum approaches a definite value: the exact area under the curve. That limit is the integral above:

Two rules make the graph method reliable:

• Area below the x-axis counts as negative (for motion along +x). Where the force points against the displacement, it does negative work. Add the areas with their signs. If x decreases instead, integrate from x_i to x_f and the signs take care of themselves.
• Use simple shapes when you can. A graph made of straight lines splits into rectangles, triangles and trapeziums, so no calculus is needed.

Work as an integral

When the force is given as a formula, integrate it. For a force that is a power of x:

for n ≠ −1, on an interval where the power is real and the integral exists. Do not apply endpoint substitution across a singularity without checking convergence. Integrate a sum term by term, and take constant factors outside the integral. The work is the value of the result at x_f minus its value at x_i.

For example (see the potential energy of a spring for the full working), the force needed to stretch a spring slowly is F = kx when the stretch is x, so the work done by this stretching force in stretching it from 0 to x_m is

which is the area of the triangle under the straight-line graph. The spring’s own force, −kx, does $-\tfrac{1}{2}kx_m^{2}$ over the same stretch. Stretching from x₁ to x₂ needs $\tfrac{1}{2}k(x_2^{2} - x_1^{2})$⁠, not $\tfrac{1}{2}k(x_2 - x_1)^{2}$⁠.

In two dimensions, the work of a variable force along a path is the sum of its component contributions, with both integrals evaluated along that same actual path:

For example, the force $(2x\,\hat i + 3y^{2}\,\hat j)$ N does 1 + 1 = 2 J of work as a body moves from (0, 0) to (1, 1) m. In this example each component depends only on its own coordinate, so the result is path-independent; that is not true for every two-dimensional force.

Where variable forces appear

• Springs and elastic cords. The force grows with the stretch, so the work done is the area of a triangle, not a rectangle.
• Pushing when tired. NCERT Example 5.5 (page 76) has a woman push a trunk with 100 N for 10 m, after which her push falls steadily to 50 N over the next 10 m. Her work is a rectangle, 100 × 10 or 1000 J, plus a trapezium, $\tfrac{1}{2}(100 + 50) \times 10$ or 750 J, in all 1750 J. Friction of 50 N, drawn on the negative side of the force axis, does (−50) × 20, that is −1000 J, over the 20 m.
• Gravity far from the Earth. The pull of gravity weakens with distance, so the work done in launching a satellite needs an integral.
• Engines and gas in a cylinder. The push of the gas on a piston changes as it expands, and the work is the area under the pressure–volume graph.

Solved examples

Common mistakes to avoid

Quick revision summary

• For a force that changes with position, split the motion into small steps over which the force is nearly constant.
• Each step contributes F(x) Δx; the total work is the limit of the sum, the definite integral of F over x.
• For motion along +x, work is the signed area under the force–displacement curve, with area below the axis negative; reverse the limits when x decreases.
• Straight-line graphs split into rectangles, triangles and trapeziums; formulas such as F = kx are integrated.

Questions students ask

What is the difference between the work done by a constant force and by a variable force?

A constant force does W = Fd cos θ, the area of a rectangle of height F cos θ and width d on the force–displacement graph. When the force changes with position, the work is found by adding up F Δx over many small steps:

This is the signed area under the force–position curve. A constant force is just the special case of a horizontal line on that graph.

Is the work done by a variable force always positive?

No, the work done by a variable force can be negative or zero. At each point the force does positive work if it points along the displacement and negative work if it points against it, so for motion along +x the area below the x-axis of the force–displacement graph counts as negative. For example, a spring’s own force does −½kx^² of work while the spring is stretched by x from its natural length.

How much work is needed to pull a hanging chain back onto a table?

Pulling a uniform chain of mass M and length L slowly back onto a smooth table, when a length L/n hangs over the edge, needs work W = MgL/(2n^²). The force needed is the weight of the part still hanging, (M/L)gy when a length y hangs, so it falls as the chain comes up, giving

for the total work. The same answer follows from raising the hanging part, of mass M/n, through L/(2n), the depth of its centre of mass.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A force F = (10 + 0.5x) N acts on a body as it moves from x = 0 to x = 4 m. Find the work done.

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   $$\begin{aligned} W &= \int_0^{4} (10 + 0.5x)\,dx \\ &= \Big[10x + 0.25x^{2}\Big]_0^{4} \\ &= 40 + 4 = 44\ \text{J} \end{aligned}$$

2. Q2. A force rises linearly with position from 10 N to 30 N while a body moves 5 m along it. Find the work done.

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   The graph is a trapezium:

   $$\begin{aligned} W &= \tfrac{1}{2}(10 + 30) \times 5 \\ &= 100\ \text{J} \end{aligned}$$

3. Q3. A force F = 2/x² N acts along the x-axis. Find the work it does from x = 1 m to x = 2 m.

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   $$\begin{aligned} W &= \int_1^{2} \frac{2}{x^{2}}\,dx \\ &= \Big[-\frac{2}{x}\Big]_1^{2} \\ &= -1 + 2 = 1\ \text{J} \end{aligned}$$