A block is kept on a smooth inclined plane of the angle of inclination 30° that moves with a constant acceleration so that the block does not slide relative to the inclined plane. Let F1 be the contact force between the block and the plane. Now the inclined plane stops and let F2 be the contact force between the two in this case. Then (F1/F2) is

Question

A block is kept on a smooth inclined plane of the angle of inclination 30° that moves with a constant acceleration so that the block does not slide relative to the inclined plane. Let F1 be the contact force between the block and the plane. Now the inclined plane stops and let F2 be the contact force between the two in this case. Then (F1/F2) is (A) 1 (B) (4/3) (C) 2 (D) (3/2)

Tardigrade Admin · Accepted Answer

When block does not slide with respect to the inclined plane,

F_{1} =

normal reaction

= m g cos θ + ma s in θ

∵ a = g t an θ

∴ F_{1} = m g cos θ + m (g t an θ) s in θ

= m g sec θ = m g sec 3 0^{\circ} = \frac{2 m g}{3}

When incline plane is stationary,

F_{2} = m g cos θ = m g cos 3 0^{\circ} = \frac{3}{2} m g

∴ \frac{F _{1}}{F _{2}} = \frac{4}{3}

$1$

$\frac{4}{3}$

$2$

$\frac{3}{2}$

1

34​

2

23​

When block does not slide with respect to the inclined plane, F1​= normal reaction =mgcosθ+masin⁡θ ∵a=gtanθ ∴F1​=mgcosθ+m(gtan⁡θ)sin⁡θ =mgsecθ=mgsec⁡30∘=3​2mg​ When incline plane is stationary, F2​=mgcosθ=mgcos⁡30∘=23​​mg ∴F2​F1​​=34​

$1$

$\frac{4}{3}$

$2$

$\frac{3}{2}$