A body is kept on a horizontal rough plane. The plane is then gradually raised to an inclination of `30^(@)` with the horizontal and the body starts to slide down. The body descends 12 m along the inclined plane in the next 4 s. Find the coefficients of static and kinetic friction between the surfaces in contact.

Explanation

In this case angle of repose `30^(@)` .
`:.` Coefficient of static friction `mu= tan 30^(@) = (1)/(sqrt(3))= 0.577`
As the body begins to slide down kinetic friction starts acting against the motion. Downward acceleration,
a = g (sin `theta-mu. "cos"theta`)
[ `mu.` = coefficient of kinetic friction ]
= 9.8 `((1)/(2)-mu.(sqrt(3))/(2))= 9.8 (0.5-0.866 mu.) "m.s"^(-2)`
As the body travels 12m in 4 s from rest from the equation,
`s = (1)/(2) at^(2), ` we get,
`12 = (1)/(2) a. (4)^(2) " ""or", a = (3)/(2) = 1.5 "m.s"^(-2)`
`:. 9.8 (0.5-0.866 mu.)`= 1.5
or , 4.9-8.5 `mu. = 1.5 " ""or", mu.= (3.4)/(8.5) `= 0.4