Question

A block of wood weighs 10 n and is resting on an inclined plane. The coefficient of friction is 0.7. The frictional force that acts on the block when the plane is 300 inclined with the horizontal is:

Answer 1

There is little mistake. Instead of 300 there will be 30 degree which means the angle of Inclination is 30°.

Weight of the Man = 10 N.

∴ Mass = 10/10 = 1 kg. [Since, g = 10 m/s².]

Coefficient of friction(μ)  = 0.7

We know in case of Incline, Normal Reaction = mgcosθ

∴ N = WCosθ

      = 10 × Cos 30°

      = 5√3

Frictional force = 5√3 × 0.7  = 6.06 N.

Hope it helps.