A block rests on a rought inclined plane making an angle of `30^(@)` with horizontal. The coefficient of static friction between the block and inclined plane is `0.8` . If the frictional foerce on the block is `10 N`, the mass of the block in `kg` is `(g = 10 m//s^(2))`

Options

A. `2.0`
B. `4.0`
C. `1.6`
D. `2.5`

Answer

A

Explanation

Since the block is resting (not moving)
`f = mg sin theta != mu_(s) mg cos theta`
`m = (f)/(g sin theta)`
` = (10)/(10 xx sin 30^(@))`
`= 2 kg`