Question

In the figure shown, a body of mass $6kg$ is at a distance of $16m$ from the pulley at $t=0$ when it is released. If the string is cut after $2s$, then find the time after which the $6kg$ block will hit the pulley. Consider the surface to be smooth (Take $g=10m/s^2$)

• A. $1s$
• B. $2s$
• C. $3s$
• D. $4s$

Answer 1

The correct option is C $3s$

FBD of $4kg$ block can be drawn as

FBD of $6kg$ block :

From the FBDs, we can say that the acceleration $a$ of the system (from $t=0$ to $t=2s$) is
$a=\frac{\text{Supporting force}}{\text{Total mass}}=\frac{4g}{4+6}=4m/s^2$
Distance covered in first $2s$ can be found using the second equation of motion.
$s=ut+\frac{1}{2}a{t}^2=\frac{1}{2}a{t}^2$
(because the system was at rest initially)
$⟹s=\frac{1}{2}\times 4\times 2^2=8m$
So remaining distance to the pulley $=16-8=8m$
After $t=2s$, string is cut, so the block will move with uniform velocity, $v=u+at=0+4\times 2=8m/s$
So, time taken to cover remaining distance of $8m$ is
$t=\frac{8m}{8m/s}=1s$
Therefore, total time taken for the block to hit the pulley will be $2+1=3s$