Question

A small block of mass $m$ is placed on a plank of mass $2m$ and length $L$. A force (\F\) is applied as shown in the figure. The time taken by block $A$ to reach the end of plank $B$ is $\sqrt{\frac{xmL}{F}}$. Find $x$.

Answer 1

Find the acceleration $a_A$ of block $A$.

As per the given diagram,

$F=2T$

$T=\frac{F}{2}$

Acceleration of block $A$ is given by,
As we know,

$a=\frac{F}{m}$

$T=m{a}_A$

Therefore,

$a_A=\frac{F}{2m}$

Find the relative acceleration $\left(a_{AB}\right)$

Acceleration of block $B$ is given by,
As we know,

$a=\frac{F}{m}$

$T=2m{a}_B$

Therefore,

$a_B=\frac{F}{4m}$

Relative acceleration $\left(a_{AB}\right)$ is given by,

$a_{AB}=a_A-a_B$

$a_{AB}=\frac{F}{4m}$

Find the time taken by block $A$ to reach the end of plank $B$.

As we know, 2𝑛𝑑 equation of motion is

$s=ut+\frac{1}{2}a{t}^2$

Hence, by putting the values we get,

$\frac{1}{2}{a}_{AB}\times t^2=L$

$t=\sqrt{\frac{2L\times 4m}{F}}sec$

Therefore,

$t=\sqrt{\frac{8mL}{F}}$

Final Answer: $8$