Question

A block is kept on the floor of an elevator at rest. The elevator starts descending with acceleration of \(5\ \text{m/s}^{2}\). The displacement of the block during the first \(0.2\ \text{s}\) after the start is \(\{\)Take \(g=10\ \text{m/s}^{2}\}\)

• A. \(0.1\ \text{cm}\)
• B. \(1\ \text{cm}\)
• C. \(10\ \text{cm}\)
• D. \(1\ \text{m}\)

Hint:

Since the lift accelerates down at less than g, the block stays in contact and moves with the lift; use s = ut + (1/2)at^2 from rest.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
If the lift accelerates downward with \(a\lt g\), the floor still pushes the block (normal force \(N=m(g-a)>0\)), so the block moves with the lift.

Step 2: Calculation
Starting from rest, \(s=\frac{1}{2}at^{2}=\frac{1}{2}\times 5\times(0.2)^{2}=0.1\ \text{m}\).

Step 3: Compare the options
\(0.1\ \text{m}=10\ \text{cm}\), which matches the third option.

Final Answer:
The block is displaced by \(10\ \text{cm}\) in the first \(0.2\ \text{s}\). \[ \boxed{\text{(C) }10\ \text{cm}} \]