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

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/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
$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ =0+\frac{1}{2}10\times {\left(0.2\right)}^2=5\times 0.04 \\ =0.2\mathrm{m}=20\mathrm{cm} \end{gathered}$$
The displacement of the body is 20 cm during the first 0.2 s.