Physics - NLM Question with Solution | TestHub

While the incline accelerates horizontally with acceleration a, the block is stationary relative to it, so in the ground frame the block also has acceleration a. Forces on the block: weight mg down, normal N perpendicular to the plane.

Resolving perpendicular to the plane (no acceleration component perpendicular to the plane in the plane's frame):

$${\mathrm{N}}_{\text{moving }}=\mathrm{mg}\cos \theta +\mathrm{ma}\sin \theta$$

with a = g tan$\theta$ for no sliding, this gives

$${\mathrm{N}}_{\text{moving }}=\mathrm{mg}\cos \theta +\mathrm{mg}\tan \theta \sin \theta =\frac{\mathrm{mg}}{\cos \theta }$$

When the plane stops, the block simply rests (or slides) on a stationary incline:

$${\mathrm{N}}_{\text{stopped }}=\mathrm{mg}\cos \theta$$

$$\frac{{\mathrm{N}}_{\text{stopped }}}{{\mathrm{N}}_{\text{moving }}}={\cos }^2\theta$$

The normal force changes by a factor of cos^2$\theta$. Correct option: (C).