Question

A wooden block of mass $8kg$ slides down an inclined plane of inclination ${30}^0$ to the horizontal with constant acceleration $0.4m/s^2$. The force of friction between the block and inclined plane is :$(g=10m/s^2)$

• A. $12.2N$
• B. $24.4N$
• C. $36.8N$
• D. $48.8N$

Answer 1

Answer: (C)

$36.8N$

Here, $mg\sin \theta -f=ma$
$mg\sin \theta -ma=f$
$8\times 10\sin \left(30\right)-8\times 0.4=f$
$40-3.2=f\Rightarrow f=36.8N$