Q.Explain with graphs the difference between work done by a constant force and by a variable force.

Step 1. Constant force. When $F$ and $\theta$ do not change during the motion, the small work $dW=(F\cos\theta)\,dr$ integrates trivially: $W=(F\cos\theta)\displaystyle\int_{r_i}^{r_f}dr=(F\cos\theta)(r_f-r_i)$.

Step 2. Graphically, plotting $F\cos\theta$ (vertical axis) against $r$ (horizontal axis) for a constant force gives a horizontal straight line. The work done between $r_i$ and $r_f$ is exactly the rectangular area under this line -- base $(r_f-r_i)$, height $F\cos\theta$.

Step 3. Variable force. When $F$ or $\theta$ (or both) change with position, $F\cos\theta$ is itself a function of $r$, so the work must be found by integrating: $W=\displaystyle\int_{r_i}^{r_f}F\cos\theta\,dr$.

Step 4. Graphically, this is the area under a curved (non-horizontal) $F\cos\theta$-versus-$r$ graph -- the shape of the curve depends on how the force varies. A stretched spring (Hooke's law force $F=kx$, growing linearly with $x$) is the standard example: its $F$-versus-$x$ graph is a straight line through the origin (not horizontal), and the work done stretching it to $x$ is the triangular area under that line, $\tfrac12kx^2$ -- illustrating how a variable force's graph, and hence the shape of the area representing work, differs fundamentally from the constant-force rectangle.

Draw and contrast the two graphs: a horizontal line (rectangle area) for constant force versus a sloped/curved line (integral area) for variable force, using the spring as the standard variable-force example.

• Using multiplication (W = F.d) for a variable force instead of integrating.
• Forgetting that the area interpretation still applies to a variable force, just with a non-rectangular shape.