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

The free body diagram of the system is shown below: nbsp; nbsp; nbsp; nbsp; The two bodies are separated because the elevator is moving downward wit ...

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Standard XII

Physics

Throwing and Catching at Same Height

A block is ke...

Question

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.

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Solution

The free-body diagram of the system is shown below:

The two bodies are separated because the elevator is moving downward with an acceleration of 12 m/s² (>g) and the body moves with acceleration, g = 10 m/s² [Freely falling body]

Now, for the block:
g = 10 m/s², u = 0, t = 0.2 s
So, the distance travelled by the block is given by
$s = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} 10 \times {(0.2)}^{2} = 5 \times 0.04 = 0.2 m = 20 cm$
The displacement of the body is 20 cm during the first 0.2 s.

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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.