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IIT JEE > IIT JEE Questions > 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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NEET,PUC,IIT Physics, .Net Trainer

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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Hi Srijin 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 with respect to earth. But the book is inside the elevator, it should be solve... read more
Hi Srijin 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 with respect to earth. But the book is inside the elevator, it should be solve in reference to elevator. The acceleration with respect to elevator is 12-9.8=2.2 m/sec^2 we know s=ut+1/2at^2 where u=0 and a=2.2 m/sec^2 s=1/2 * 2.2* 0.2^2 s=0.04m above the elevator floor. read less
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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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Problem Cracker

As the acceleration of the elevator is greater than gravitational acceleration, the lift will freely fall under gravity. So, the displacement of the bidy will be S=0.5*9.8*0.2*0.2=0.196 m.
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IIT Foundation Expert

s=Ut+(1/2)at^2 Here U=0, t=0.2 Sec, and a=-12 m/s^2 {-ve sign is because its descending object} So Displacement S = (0*0.2) + 1/2 (-12*0.2*0.2) =-0.24 m
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