JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Laws of Motion

Constant velocity on a frictionless surface means zero net force, so the applied force balances the along-track component of gravity at every point. On the ascending slope the component $mg\sin \theta =2\text{N}$ acts backward, so the applied force is $+2\text{N}$. On the descending slope this component acts forward, so the applied force reverses to $-2\text{N}$. The applied force is $+2\text{N}$ over the first half of the horizontal span and $-2\text{N}$ over the second half, with equal step widths.