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 N
B. 24.4 N
C. 36.8 N
D. 48.8 N

Answer

C

Explanation

Here mg `sin theta - f = ma `

mg `sin theta - ma = t`
`8 xx 10 sin (30^(@)) - 8 xx 0.4 = t`
`40 - 3.2 = f implies f = 36.8 N`