A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s 2 . Find the displacement of the block during the first 0.2 s after the start. Take g = 10 m/s 2 .

Question

Philoid · Accepted 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

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.

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.