Question

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

• A. $36.8\text{N}$
• B. $76.8\text{N}$
• C. $65.6\text{N}$
• D. $\text{None of these}$

Answer 1

The correct option is A $36.8\text{N}$


$mg\sin \theta -f=ma$
Where $f=$Force of friction.

or $f=m(g\text{sin}\theta -a)$
$=8(10\times \frac{1}{2}-0.4)=36.8\text{N}$