A wooden box of mass `8kg` slides down an inclined plane of inclination `30^(@)` to the horizontal with a constant acceleration of `0.4ms^(-2)` What is the force of friction between the box and inclined plane ? `(g = 10m//s^(2))` .

Options

A. 12.2 BN
B. 24.4 N
C. 36.8 N
D. 48.8 N

Answer

C

Explanation

( c) Here, mg `sin theta -f=ma`

`mg sin theta-ma =f`
`implies 8xx10 sin30^(@)-8xx0.4=f`
`implies 40-3.2=f`
`implies f=36.8 N`