A box of mass m = 10 kg resting on a frictionless horizontal ground is acted upon by a horizontal force F, which varies as shown. Find speed of the particle (in m/s) when the force ceases to act.

The area under force-time graph gives the impulse. Hence impulse acting on the box.J→=∫titf F→dt= Area under the graph =12×100×2=100N−sUsing impulse-momentum equation:mv→f=mv→i+∫titf F→dt or mv→f=mv→i+J→10×v=10×0+100 ⇒ v=10m/s