Question

A plank of mass $2\text{kg}$ and length $9\text{m}$ is placed on a horizontal floor. A small block of mass $1\text{kg}$ is placed on top of the plank at its right extreme end. The coefficient of friction between the plank and the floor is $0.5$ and that between the plank and the block is $0.2$. If a horizontal force of $30\text{N}$ starts acting on the plank to the right, the time after which the block will fall off the plank is
(Take $g=10{\text{ms}}^{-2}$)

• A. $2.96\text{s}$
• B. $2\text{s}$
• C. $1.75\text{s}$
• D. $3.14\text{s}$

Answer 1

The correct option is B $2\text{s}$

Let the given condition be as shown below

Let $f_1$ be the maximum friction acting between block and plank and $f_2$ be the maximum friction acting between plank and floor. So we have FBD of both block and plank as
Therefore, we have
$f_1=\mu N_1=0.2\times 10=2\text{N}$
Similarly, $f_2=\mu N_2=0.5\times 3\times 10=15\text{N}$
Thus we have,
$a_1=\frac{f_1}{1}=2{\text{m/s}}^2$
and, $a_2=\frac{30-(f_1+f_2)}{2}=\frac{30-17}{2}=6.5{\text{m/s}}^2$
Also, relative acceleration between the block and plank is given as
$a_r=6.5-2=4.5{\text{m/s}}^2$
Hence, time taken by the block to fall off the plank is given by
$s_r=u_{r}t+\frac{1}{2}{a}_r{t}^2\Rightarrow t=\sqrt{\frac{2s_r}{a_r}}$ (since, $u_r=0$)
$\Rightarrow t=\sqrt{\frac{2\times 9}{4.5}}=2\text{sec}$