Sum

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}.

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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/s^{2} (>g) and the body moves with acceleration, *g* = 10 m/s^{2 }[Freely falling body]

Now, for the block:*g* = 10 m/s^{2}*, u* = 0*, t* = 0.2 s

So, the distance travelled by the block is given by

\[s = ut + \frac{1}{2}a t^2 \]

\[ = 0 + \frac{1}{2}10 \times \left( 0 . 2 \right)^2 = 5 \times 0 . 04\]

= 0 . 2 m = 20 cm

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

Concept: Newton’s Second Law of Motion

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