Question

A block of mass $5\text{kg}$ is resting on a rough horizontal surface. Now a force of $24N$ is imparted to it with a negligible impulse as shown in the figure. If the coefficient of kinetic friction is 0.4 and $g=10{\text{m/s}}^2$, then the acceleration of the block will be

• A. $0.26m/s^2$
• B. $0.39m/s^2$
• C. $0.69m/s^2$
• D. $0.8m/s^2$

Answer 1

The correct option is D $0.8m/s^2$

The FBD can be shown as given below.

Given, coefficient of kinetic friction ${\mu }_k=0.4$
$\therefore$ we have friction force
$f={\mu }_k.N={\mu }_k.mg=0.4\times 5\times 10N=20N$

Hence, acceleration of the mass is
$a=\frac{F_{net}}{m}=\frac{24-f}{m}=\frac{24-20}{5}=0.8m/s^2$