A coin slides down an inclined plane of inclination `phi` at a constant speed. Prove that if the coin is pushed up with a velocity u on that plane it can rise up to ` (u^(2))/(4 g "sin"phi)` .

Explanation

Since the coin slides down Fig with a uniform speed it has no acceleration along the plane. So,
`mu.R= mg sin phi`
Downward force acting on the coin as it is pushed up,
F = mg sin `phi+mu. R = mg sin phi +mg sin phi = 2 mg sin phi`
Retardation , a = `(2mg sin phi)/(m) = 2 g sin phi`.

If the coin moves up to s then,
`u^(2) = 2as or , s = (u^(2))/(2(2g sin phi)) = (u^(2))/(4 g sin phi)`.
As the coin stops and attempts to come down limiting friction acts on it. Downward force mg sin `phi` along the plane is equal to `mu.` R which is always less less than `mu`R , as `mu gt mu.` (since Hence the coin cannot slide down again.