A block of mass `m` sides down an inclined right angled trough .If the coefficient of friction between block and the trough is `mu_(k)` acceleration of the block down the plane is

Options

A. `g ( sin theta + sqrt2 mu_(k) cos theta)`
B. `g ( sin theta + mu_(k) cos theta)`
C. `g ( sin theta - sqrt2 mu_(k) cos theta)`
D. `g ( sin theta - mu_(k) cos theta)`

Answer

C

Explanation

Resultant of `N` and `N (= sqrt2 N)` is equal to `mg cos theta`
` :. sqrt2 N = mg cos theta`
`:. N = (mg cos theta)/(sqrt2)`

Now kinetic friction will act from two sides
`:. a= (mg cos theta - 2 mu_(k) N)/(m)`
Subtituting the value of `N`, we get
`a = g (sin theta - sqrt2 mu_(k) cos theta)`