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 fiction 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

Given : $m=8$ kg $a=0.4m/s^2$

Equation of motion of block : $ma=mg\sin {30}^o-f$

$\therefore$ $8\times 0.4=8\times 10\times 0.5-f$

$⟹$ $f=40-3.2=36.8$ $N$