Question

A stationary body of mass $m$ is slowly lowered onto a rough massive platform moving at a constant velocity $v_0=4\text{m/s}$. If the displacement of the body with respect to the platform $(\mu =0.2)$ is $x\text{m}$, then $x$ is
(Take $g=10{\text{m/s}}^2$)

• A. $4$
• B. $6$
• C. $12$
• D. $8$

Answer 1

The correct option is A $4$


With respect to platform, the initial velocity of the body of mass $m$ is $4\text{m/s}$ towards left (opposite to massive platform) and it starts retarding at the rate of $a$ due to friction. So,

$a=\frac{f}{m}=\frac{\mu mg}{m}=\mu g$

$a=0.2\times 10=2{\text{m/s}}^2$

Using equation of motion,

$v^2=u^2+2as$

$\Rightarrow 0=4^2+(2\times -2\times x)(\because a=-2{\text{m/s}}^2)$

$\Rightarrow x=4\text{m}$

Hence, option $\left(a\right)$ is correct.