Question

A wooden block of mass $8$kg slides down an inclined plane of inclination ${30}^o$ 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. $12.2$N
• B. $24.4$N
• C. $36.8$N
• D. $48.8$N

Answer 1

Answer: (C)

$36.8$N

Mass of the block $m=8$ kg

Acceleration of the block $a=0.4m/s^2$

Since the block is sliding down the plane, so friction force acts backwards as shown.

Equation of motion along the inclined plane :

$ma=mg\sin {30}^o-f$

We get $f=mg\sin {30}^o-ma$

$\therefore$ $f=\left(8\right)\left(10\right)\left(0.5\right)-\left(8\right)\left(0.4\right)$

$⟹f=36.8$ N