Question

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

Answer 1

Because the elevator is moving downward with an acceleration $12m/s^2$ (> g), the body gets separated.

So, body moves with acceleration,

g = 10 $m/s^2$ [Freely falling body]

and the elevator moves with acceleration $12m/s^2$.

Now, the bock had acceleration,

g = 10 $m/s^2$

u = 0

t = 0.2 sec

So, the distance travelled by the block is given by,

$S=ut+\frac{1}{2}a{t}^2$

$=0+\frac{1}{2}10(0.2{)}^2=5\times 0.04$

= 0.2 m = 20 cm

The displacement of body is 20cm during first 0.2 sec.