A block of mass m is sliding down a stationary inclined plane. The base of the inclined plane has a length l, and the coefficient of kinetic friction between the block and inclined surface is 0.14. What should be the inclination of the plane so that the block can slide down to the ground in minimum time?

A block of mass m is sliding down a stationary inclined plane. The bas

1.	A block of mass m is sliding down a stationary inclined plane. The base of the inclined plane has a length l, and the coefficient of kinetic friction between the block and inclined surface is 0.14. What should be the inclination of the plane so that the block can slide down to the ground in minimum time?
Answer» Solution :Acceleration of the block ALONG the plane , a = g (sin `theta -mu. "cos" theta` ) Let the time required to slide down from A to B be t Fig. `:.AB = (1)/(2)at^(2)= (l)/("cos"theta)` or, `(1)/(2)t^(2)g (sintheta -mu."cos"theta) = (l)/("cos"theta)` or, `t^(2) = (2L)/(g) (1)/("sin"theta "cos"theta-mu."cos"^(2)theta)` `=(4l)/(g)[(1)/(sin 2thetacostheta-mu.(1+cos^(2)theta))]` When t is MINIMUM `t^(2)`is also minimum. `:. sin 2thetai-mu (1+cos2 theta)` = x (say)is maximum. `:. (dx)/(d theta)= 0 " ""or", 2 cos2 theta -mu. (0-2 SIN2 theta)` =0 or, cos 2 `theta +mu sin2 theta` =0 or , `tan2theta = -(1)/(mu.)=-(1)/(0.14) = -tan 82^(@)` or, `tan 2 theta = tan(180^(@) -82^(@)) = tan 98^(@)` `:. 2theta= 98^(@) "and"theta = 49^(@)`.