Question

A block of mass, $M=5kg$ is resting on a rough horizontal surface for which the coefficient of friction is $0.2$ . Let a force, $F=40N$ be applied, what will then the acceleration of the block be?

• A. $5.73m{s}^{-2}$
• B. $8.0m{s}^{-2}$
• C. $3.17m{s}^{-2}$
• D. $10.0m{s}^{-2}$

Answer 1

Answer: (A)

FBD of the block is shown below

Kinetic friction acting on the body $={\left(\mu \right)}_{k}R=0.2\left(Mg-Fsin{\left(30\right)}^o\right)$
$=0.2\left(5\times 10-40\times \frac{1}{2}\right)=0.2\left(50-20\right)=6N$

By Newton's second law of motion, acceleration of the block
$=\frac{netforce}{mass}$ $=\frac{Fcos30^o-kineticfriction}{mass}$ $=\frac{40\times \frac{\sqrt{3}}{2}-6}{5}=5.73m/s^2$