Question

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 sec after the start. take g=10

Answer 1

@Vincent-Lyon.Fr sir

Answer 2

Do you mean 1.2 m/s² ?

Answer 3

no 12 m/s^2

Answer 4

It means the elevator is going faster than free fall. So the object would simply fall under gravity.

Answer 5

so \[s=\frac{1}{2}g (0.2)^2\]\[s=5 \times 0.04=0.2m\]

Answer 6

If the elevator had descended with an acceleration of 8 m/s^2. then wht will be the answer ??

Answer 7

Correct.
If the elevator has 8m/s², the body stays in contact with it and will go down with this acceleration.

Answer 8

thanx A lot sir :)