Q. 423.7( 11 Votes )

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

Answer :


Given:


The two bodies are separated because the elevator is moving downward with an acceleration of 12 ms-2 (>g) and the body moves with acceleration, g = 10 ms-2 [Freely falling body]


Now, for the block: g = 10 m/s �, u = 0, t = 0.2 s


So, the distance travelled by the block is given by



=5×0.04 =0.2 m =20 cm


The displacement of the body is 20 cm during the first 0.2 s


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