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.

Answer 1

Given :

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=ut+1/2at²

=0+1/210×(0.2)²

=5×0.04

=0.2 m

=20 cm

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

Answer 2

Answer:

Effective gravity is 10−12=−2m/s

2

The block will lose the contact with the elevator.

Thus, the acceleration of the block is 10m/s

2

and the displacement of the block in first 0.2 s is

2

1

×10×(0.2)

2

=0.2m=20cm