Q.A uniform force of (2i^ + j^) N acts on a particle of mass 1 kg. The particle displaces from position (3j^ + k^) m to (5i^ + 3j^) m. The workdone by the force on the particle is :

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

$W = (F\cos\theta)(r_f - r_i)$ …