Question

A plank of mass $2\text{kg}$ and length $9.5\text{m}$ 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 plank and floor is $0.5$ and that between plank and 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 $2\text{s}$

Limiting friction between plank and ground is:
$F_1={\mu }_1{N}_1$ ...(i)
Limiting friction between block and plank is:-
$F_2={\mu }_2{N}_2$ ...(ii)

FBD of plank + block:-


Balancing vertical forces:
$N_1=30\text{N}$

For block of $1\text{kg}$:-


$N_2=1g=10\text{N}$

Since relative motion is happening between block and plank, kinetic friction $F_2$ will act on block.
$\Rightarrow F_2=m_2{a}_2$ ....(iii)
and $F_2={\mu }_2{N}_2=0.2\times 10=2\text{N}$

FBD of plank (Horizontal forces only):


$30-F_1=m_1{a}_1$
and $F_1={\mu }_1{N}_1=0.5\times 30=15\text{N}$
$\therefore a_1=\frac{30-15}{2}=7.5{\text{m/s}}^2$

Similarly, for block:
$F_2=m_2{a}_2$
$\Rightarrow a_2=\frac{2}{1}=2{\text{m/s}}^2$

$a_{rel}=a_{21}=a_2-a_1=-2-(+7.5)=-9.5{\text{m/s}}^2$
(Relative acceleration of block will be towards -ve $x$ axis)

Applying 2nd equation of motion :-
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$\Rightarrow -9.5=0-\frac{1}{2}\left(9.5\right)t^2$
$\therefore t^2=2$
$\Rightarrow t=\sqrt{2}=1.41\text{s}$