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IIT JEE > A block is kept on the floor of an elevator at rest....
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A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration 12 m/s. Find the displacement of the block during the first 0.2 s after the start?

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Since the acceleration of elevator is greater than g (Acceleration due to gravity 9.8 m/s2 ) and block and elevator are not attached, the block will get separated and move like free fall body with acceleration g. we know s=ut+1/2at^2 where u=0 and a=g(free falll acceleration) s=1/2 * 10* 0.2^2...
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English,maths,history, environmental science ,geography,civics, untill primary level

As in the given question acceleration is more than the gravitational pull 'g' i.e., a=12>g=9.8m/s^2, this easily shows its a freely falling body such that a=g, u=0 (given at rest), t=0.2s (given). So substitute this in s=ut+(1/2)a(t^2) ad solve.
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Physics, IIT expert Faculty at MVN sector-17 Faridabad

Acceleration should be 12m/s^2. If such and block is not bound to be with the elevator then it will go down with acceleration of 'g'. Then displacement in first 0.2s will be s=1/2gt^2 (taking u=0). S=0.2 m
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check it
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check the unit of acceleration. Since , elevator descends with acceleration greater than that of acceleration due to gravity, the block will lose contact from the floor. In the ground frame , acceleration of block will be g=10m/s^2 downwards, so in 0.2 second displacement of block=(1/2)*(10)*(.2)^2=0.2...
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check the unit of acceleration. Since , elevator descends with acceleration greater than that of acceleration due to gravity, the block will lose contact from the floor. In the ground frame , acceleration of block will be g=10m/s^2 downwards, so in 0.2 second displacement of block=(1/2)*(10)*(.2)^2=0.2 m =20 cm downwards In the elevator frame, acceleration of the block will be 2 m/s^2 upwards. So, displacement will be =(1/2)*(2)*(.2)^2=0.04 m =4 cm upwards read less
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English,maths,history, environmental science ,geography,civics, untill primary level

As it is freely falling body because given acceleration a=12>g=9.8 m/s^2, so u=0,a=g=9.8,t=0.2s. So the required formula is s=ut+(1/2)at^2 => s=0.196 m is the required solution.
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Physics Tutor

You need to understand the concept involved here. One thing is attached to another body which is coming down only till the time the falling acceleration is less than the value of "g".Now that comes with the equation of N-mg = ma which I believe you must be aware of.Here the acceleration of elevator is...
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You need to understand the concept involved here. One thing is attached to another body which is coming down only till the time the falling acceleration is less than the value of "g".Now that comes with the equation of N-mg = ma which I believe you must be aware of.Here the acceleration of elevator is more than the acceleration due to gravity, that would result in block getting detached from elevator floor as soon as elevator starts. So this became a very simple question of block falling freely under gravity. That means 1/2gt^2 will give you answer :) read less
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Physics by Ashwani Sir

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I have 2 years of experience

d = u*t + 0.5*a*t^2 u=0 then d = 0.5*12*0.2*0.2 d=.24
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