Q.A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of $100\ \text{N}$ over a distance of $10\ \text{m}$. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to $50\ \text{N}$. The total distance through which the trunk has been moved is $20\ \text{m}$. Plot the force applied by the woman and the frictional force, which is $50\ \text{N}$, versus displacement. Calculate the work done by the two forces over $20\ \text{m}$.

Concept: Work Energy Theorem — the net work done by all forces equals the change in kinetic energy. Here we directly compute work from force-displacement graphs.

Step 1: Force applied by woman

From $0$ to $10\ \text{m}$, force is constant at $100\ \text{N}$.

From $10$ to $20\ \text{m}$, force decreases linearly from $100\ \text{N}$ to $50\ \text{N}$.

Step 2: Work done by woman

Work = area under force-displacement graph.

• First segment (rectangle): $100 \times 10 = 1000\ \text{J}$
• Second segment (trapezium): $\frac{100+50}{2} \times 10 = 750\ \text{J}$ Total work by woman = $1000 + 750 = 1750\ \text{J}$ …

The figure is a force–displacement graph. The horizontal axis is displacement $x$ in metres; the vertical axis is force $F$ in newtons. Two curves are plotted. A solid blue line shows the force $F$ applied by the woman. From point B to point C (displacement 0 to 10 m) this force is constant at 100 N. From C to E (10 m to 20 m) it falls linearly to 50 N. A dashed line shows the opposing frictional force $f$, which is constant at $-50$ N from point G to point H (the full 0 to 20 m displacement). The negative sign means friction acts opposite to the direction of displacement. A rectangle ABCD is shaded, with A and D on the displacement axis at 0 and 10 m, and B and C at the 100 N level. Points A through I label key positions on the curves and axes.

The physical idea is central: when a force varies with position, the work done is not simply force times displacement. Instead, work equals the area under the force–displacement curve. For the applied force $F$, the work done from $x_i$ to $x_f$ is

The figure makes this concrete. The work done by the woman from 0 to 10 m is the area of rectangle ABCD: $100 \text{ N} \times 10 \text{ m} = 1000 \text{ J}$. From 10 to 20 m, the force drops linearly, so the work is the area of the trapezoid under that sloping segment — the textbook calculates it as the average force times displacement, or directly from the graph. The total work by the applied force is the sum of these two areas.

The frictional force is constant at $-50$ N over the entire 20 m. Its work is simply $f \times \Delta x = (-50 \text{ N})(20 \text{ m}) = -1000 \text{ J}$, which is the area of the dashed rectangle (negative because the force opposes motion). The net work done on the object is the sum: $1000 \text{ J} + (\text{work from 10–20 m}) - 1000 \text{ J}$.

The shaded rectangle ABCD is a visual anchor: it isolates the constant-force segment so you see that even when the force later changes, the work for that part is still just the area of a rectangle. The dashed friction line at $-50$ N reinforces that opposing forces contribute negative work, and that constant opposing forces produce a rectangular area too — but below the axis.

Why the Work-Energy Theorem is the natural lens

Work is not just a formula — it is the transfer of energy by a force acting over a distance. When a force changes with position, you cannot simply multiply force and displacement; you must find the area under the force-displacement graph. That area is the work.

Here, the woman's applied force changes in two distinct phases: constant at first, then linearly decreasing. Friction, however, is constant throughout. The problem asks for the work done by each force separately — so we treat them as independent agents, each contributing its own area.

Step-by-step solution

1. Sketch the force-displacement graphs

We have two forces to plot:

• Applied force by the woman:

  From $x = 0$ to $x = 10\ \text{m}$, it is constant at $100\ \text{N}$.

  From $x = 10\ \text{m}$ to $x = 20\ \text{m}$, it drops linearly from $100\ \text{N}$ to $50\ \text{N}$.

• Frictional force:

  Constant at $50\ \text{N}$ throughout, but acting opposite to the direction of motion. On a force vs. displacement graph, we take the magnitude for area calculation, then assign the sign based on direction. …

Concept: Work as the Area Under a Force–Displacement Graph

Step 1: Split the woman's applied force into its two phases.

Constant $100\,\text{N}$ from $x=0$ to $10\,\text{m}$; linearly decreasing from $100\,\text{N}$ to $50\,\text{N}$ from $x=10$ to $20\,\text{m}$.

Step 2: Work done by the woman = sum of the two areas.

Rectangle: $W_1 = 100 \times 10 = 1000\,\text{J}$.

Trapezoid: $W_2 = \dfrac{100+50}{2}\times 10 = 750\,\text{J}$.

$W_{\text{woman}} = 1000 + 750 = 1750\,\text{J}$ …