The upper half of an inclined surface is perfectly smooth but the lower half is rough . A body starts sliding down the plane and stops immediately on reaching the bottom. The inclination of the plane is `30^(@)` with the horizontal. Show that the frictional resistance of the rough part of the surface is equal to the weight of the body.

Explanation

Let the length of the inclined plane be 2l , and the vertical height from the ground be h. Then we get,
`sin 30^(@) = (h)/(2 l) " ""or", (1)/(2) = (h)/(2l) " ""or", h= l`.
Potential eneragy of the body of mass m at maximum height on the plane = mgh = mgl. The body comes to rest at the bottom. Hence all its potential energy at the top is spent in doing work against the force of friction which acts along teh length of the lower half of the surface. If f is the frictional force then
`fxx l ` mgl
or, f = mg = weight of the body (Proved).