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  • A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 15m/s^2.

Speed of block at the end of one second after the start w.r.t.ground is(g=10m/s^2.

(We have to find this)

Answers (1)

best_answer

We have,
Acceleration of the elevator = 15 m/s^2 ( greater than g )

Therefore, the body gets separated from the elevator. Thus body moves with the acceleration of 10 m/s^2
Initially block is at rest 
therefore,  u = 0
                  t = 0.2 sec
So, the speed of the block =  v = u + at
                                         =0 + 10 \times 0.2 = 2m/s^2

 

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manish

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