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)
A. 30 cm
B. zero
C. 20 cm
D. 25 cm

Answer 1

Here initially the block was in contact with the elevator but when the elevator moves downwards with an acceleration ‘a’, then the net acceleration of the elevator is given by $=a-g$.Since, the acceleration of the elevator is greater than the value of acceleration due to gravity, the block will gets detached from the floor of the lift. So, the block moves with acceleration $g=10m/{{s}^{2}}$.

Complete step by step answer:
Now the block has the acceleration, $g=10m/{{s}^{2}}$. We know that the equation for displacement for a body in uniform acceleration is given as, $s=ut+\dfrac{1}{2}a{{t}^{2}}$, ‘u’ is the initial speed, ‘a’ is the acceleration and ‘t’ is the time.
Given time, $t=0.2s$ and initial speed of block is, $u=0m/s$.
Substituting the values in the equation we get,
$s=0\times 0.2+\dfrac{1}{2}\times 10\times 0.{{2}^{2}}\\\ \Rightarrow s=0.2m\\\ \therefore s=20cm$
Therefore, the displacement of the body for the time period 0.2 second is 20cm.

So, the correct option is C.

Note: The most important point here is to find the net acceleration due to which the body starts to lose contact. After losing contact the body can be assumed to have an acceleration due to gravity and therefore the displacement is $s=ut+\dfrac{1}{2}a{{t}^{2}}$, ‘u’ is the initial speed, ‘a’ is the acceleration equal to g and ‘t’ is the time. When the lift moves upwards then the acceleration of the elevator is given by $=g+a$