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Free Body Diagrams

In the figure...

Question

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

1 s

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2 s

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3 s

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4 s

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Solution

The correct option is C $3 s$
FBD of $4 k g$ block can be drawn as

FBD of $6 k g$ block :

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

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In the figure shown, a body of mass 6 kg is at a distance of 16 m from the pulley at t=0 when it is released. If the string is cut after 2 s, then find the time after which the 6 kg block will hit the pulley. Consider the surface to be smooth (Take g=10 m/s2)

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