A block of mass 2 kg is lying on an inclined plane, inclined to the horizontal at 30° . If the coefficient of friction between the block and the plane is 0.7, then magnitude of frictional force acting on the block will be

Question

A block of mass 2 kg is lying on an inclined plane, inclined to the horizontal at 30° . If the coefficient of friction between the block and the plane is 0.7, then magnitude of frictional force acting on the block will be (A) 11.9 N (B) 1.19 N (C) 0.19 N (D) 11.0 N

Tardigrade Admin · Accepted Answer

The various forces acting on the block are as shown Frictional force

= μ R

... (i)
where

μ

is coefficient of friction and

R

the normal reaction of the surface on the block.
Also,

R = m g cos 3 0^{\circ}

... (ii)
From Eqs. (i) and (ii),
we get

F = μ m g cos 3 0^{\circ}

Given,

μ = 0.7, m = 2 k g, g = 9.8 m / s^{2}

,

cos 3 0^{\circ} = \frac{3}{2}

∴ F = 0.7 \times 2 \times 9.8 \times \frac{3}{2}

= 11.9 N

Q. A block of mass 2kg is lying on an inclined plane, inclined to the horizontal at 30∘ . If the coefficient of friction between the block and the plane is 0.7, then magnitude of frictional force acting on the block will be

11.9 N

1.19 N

0.19 N

11.0 N

Q. A block of mass $2 k g$ is lying on an inclined plane, inclined to the horizontal at $3 0^{\circ}$ . If the coefficient of friction between the block and the plane is $0.7,$ then magnitude of frictional force acting on the block will be