Q.A force of $20\ \text{N}$ acts on a block, making an angle of $60^\circ$ with the horizontal, and displaces the block through $4\ \text{m}$ along the horizontal floor. Find the work done by the force.

When a force $F$ acting on a body remains constant in magnitude throughout the displacement (with a fixed angle $\theta$ between the force and the direction of motion), the small work done for a small displacement $dr$ is $dW=(F\cos\theta)\,dr$. Because $F\cos\theta$ does not change with position, integrating from an initial position $r_i$ to a final position $r_f$ gives simply

Graphically, plotting $F\cos\theta$ against $r$ gives a horizontal straight line, and the work done is the rectangular area under it between $r_i$ and $r_f$. This constant-force case covers a large fraction of textbook problems -- lifting a load vertically, sliding a block down a frictionless incline (where the constant component of gravity along the slope is $mg\sin\theta$), a weightlifter raising a barbell at constant speed -- and is also the setting in which work is most easily related to a force expressed as a fixed vector, $W=\vec F\cdot\vec d$, computed directly as a dot product between the (constant) force vector and the net displacement vector, without needing any integration at all.