Question

A block is kept on a smooth inclined plane of angle of inclination ${30}^{\circ }$ that moves with a constant acceleration so that the block does to slide relative to the inclined plane.
Let $F_1$ be the contact force between the block and the plane. Now the inclined plane stops, then let $F_2$ be the contact force between the block and the plane. The value of $\frac{F_1}{F_2}$ is

• A. $1$
• B. $\frac{4}{3}$
  
• C. $2$
• D. $\frac{3}{2}$
  

Answer 1

The correct option is B $\frac{4}{3}$


Since the surface is smooth, the force of friction between the block and the plane is zero.
So, contact force is really the normal reaction between the two.

In this first case,
$F_1\sin {30}^{\circ }=ma$
and $F_1\cos {30}^{\circ }=mg$ or $F_1=\frac{mg}{\cos {30}^{\circ }}$
and in the second case,
$F_2=mg\cos {30}^{\circ }$

$\therefore \frac{F_1}{F_2}=\frac{1}{{\cos }^2{30}^{\circ }}=\frac{4}{3}$
Hence,the correct ansewer is option (b).