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) , and from there it will not slide down again.

A coin slides down an inclined plane of inclination phi at a constant

1.	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) , and from there it will not slide down again.
Answer» Solution :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.