Question

A block is kept on a smooth inclined plane of angle of incli-
nation @, which moves with a constant acceleration so that the
block does not slide relative to the inclined plane. If the inclined
plane stops, the normal contact force offered by the plane on the
block changes by a factor
(a) cos@ (b) sec@ le) cos@ (d) none of these
​

Answer 1

Answer:

$cos^{2} \theta$

Explanation:

Since the inclined place is smooth, there will be no friction force.

Assume the mass of the block = m

In the frame of reference of the inclined plane the forces acting on the block of mass m will be as shown in the figure attached.

Let the force on the block be F

The resolution of the forces as given in figure

Since the block does not slide

∴ Fcosθ = mgsinθ

or, F = mgsinθ/cosθ

N = mgcosθ + Fsinθ

  = mgcosθ + (mgsinθ/cosθ)sinθ

  = mg/cosθ

When the inclined plane stops, the force F will be zero

Hence, new Normal reaction = mgcosθ

This is greater than previous Normal Reaction by a factor = $cos^{2} \theta$

Answer 2

Answer:

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