Question

If the coefficient of friction of a plane inclined at 45' is 0.5, then acceleration of a body sliding freely on it is

Answer 1

Let the mass of the body be m.

Force of gravity along the inclined plane (F) = mgSin 45°

Normal force by the inclined plane (N) = mgCos 45°

Friction force = μN

Coefficient of friction (μ) = 0.5

F (net) = mgSin 45° - μN

ma = mgSin 45° - μmgCos 45°

Take "m" common and cancel it on both the sides.

a = gSin 45° - μgCos 45°

a = 6.93 - 3.46

a = 3.47 m/s^2

Hence, the acceleration down the inclined plane = 3.47 m/s^2