A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration $12 m / s^{2}$ . find the displacement of the block during the first 0.2 s after the start. Take $g = 10 m / s^{2}$

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Solution

Since the acceleration of the elevator is greater than the acceleration due to growing. Hence the block will do free ball and will lose contact with the elevator.
Hence acceleration of block = $10 m / s^{2}$
$T = 0.2 s e c$
Displacemetn (S)= $\frac{1}{2} a t^{2}$
$= \frac{1}{2} \times / 10 \times \frac{/ 2}{/ 10} = \frac{/ 2}{/ 10} = \frac{1}{5} m = 0.2 m$

Since, Mg-N=ma
If $a = 12 m / s^{2}$ (If block goes with elevator)
$m (10) - N = m (12)$
N=-2m $\to$ Not possible because minimum value of N can be zero.
Hence a will be equal to gravity = $10 m / s^{2}$

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A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration 12m/s2. find the displacement of the block during the first 0.2 s after the start. Take g=10m/s2

A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration $12 m / s^{2}$ . find the displacement of the block during the first 0.2 s after the start. Take $g = 10 m / s^{2}$