Q.State the formula for calculating work done by a force. Are there any conditions or limitations in using it directly? If so, state those clearly. Is there any mathematical way out for it? Explain.

[!TLDR] $W=Fs\cos\theta$ works only for a CONSTANT force and displacement; for a variable force, use $W=\int F\,ds$ (the area under the F-s graph) instead. [!ANS …

The standard formula for work done is $W=\vec{F}\cdot\vec{s}=Fs\cos\theta$, where $\theta$ is the angle between the applied force $\vec{F}$ and the displacement $\vec{s}$. This formula has a real LIMITATION: it is valid ONLY when both the force and the displacement are CONSTANT, finite vectors throughout the motion. In many real situations (e.g. lifting an object through a large height where gravity's effective value or direction shifts, or a fluid-drag force that depends on the object's changing speed), the force VARIES during the motion, and the simple formula cannot be applied directly. The mathematical way out is INTEGRATION (4.5.5): break the total displacement into a large number of tiny (infinitesimal) steps ds, over each of which the force is practically constant, compute the elementary work $dW=F\,ds$ for each tiny step, and sum (integrate) over the whole displacement: $W=\int_{s_1}^{s_2}F\,ds$. This integral …

State the constant-force work formula, identify its stated limitation (validity only for constant F and s), and give …

Applying $W=Fs\cos\theta$ directly to a clearly variable force (e.g. a spring or a distance-dependent for …