Question

A block is kept on a smooth inclined plane whose angle of inlination is $\theta$ that 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. $\tan \theta$
• B. ${\tan }^2\theta$
• C. ${\cos }^2\theta$
• D. ${\cot }^2\theta$

Answer 1

The correct option is C ${\cos }^2\theta$

When the plane is moving

Applying pseudo force

As, the block is at rest,

$gsin\theta =acos\theta$

thus, $a=gtan\theta$
Now, $Nsin\theta =ma=mgtan\theta$
$\Rightarrow N_1=\frac{mg}{cos\theta }...........\left(i\right)$
When the plane is stationary
$N2=mgcos\theta .............\left(ii\right)$
From eq. (i ) and(ii)
$\frac{N_1}{N_2}=\frac{1}{co{s}^2\theta }$
or $N_2=N_{1}co{s}^2\theta$
Thus correct answer is (c)