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# A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 meters per square second. Find the displacement of the block during the first 0.2 seconds after the start. Take $g = 10{\text{ m/}}{{\text{s}}^2}$.

Last updated date: 19th Sep 2024
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Hint: In this question, we need to determine the displacement of the block inside the elevator during its journey in the first 0.2 seconds. For this, we will use Newton’s equations of motion.

Complete step by step answer:As the elevator is moving downwards with the acceleration higher than the acceleration due to gravity so, the coin placed on the floor of the elevator will release the surface, and consequently, now, the coin is also traveling with the acceleration due to gravity. The following figure depicts the same situation as:

The relative acceleration on the coin is $10 - 12 = - 2{\text{ m/}}{{\text{s}}^2}$ which is against the direction of the acceleration due to gravity.
Following the third Newton’s equation of motion, which is given by $S = ut + \dfrac{1}{2}a{t^2}$ where, S is the displacement of the body (here, coin), ‘u’ is the initial velocity of the body, ‘t’ is the time, and ‘a’ is the acceleration of the body.
So, substitute the initial velocity as 0, t as 0.2 seconds, and acceleration of the coin as 10 meters per square seconds in the formula $S = ut + \dfrac{1}{2}a{t^2}$ to determine the displacement of the coin inside the elevator.
$S = ut + \dfrac{1}{2}a{t^2} \\ = (0)(0.2) + \dfrac{1}{2}\left( {10} \right){\left( {0.2} \right)^2} \\ = 5(0.04) \\ = 0.2{\text{ meters}} \\ = 20{\text{ cm}} \\$
Hence, the coil will be displaced by 20 centimeters when the elevator comes downwards with an acceleration of 12 meters per square second.

Note:It is interesting to note here that as the elevator is coming downwards so, the coin leaves the surface of the elevator, but if the elevator goes upwards then, the coin will remain stick to the floor only as the relative acceleration will be in the direction of the gravitational acceleration only.