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Q. A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of $12 \,m / s ^{2}$. The displacement (in $cm$ ) of the block during the first $0.2 s$ after the elevator starts descending is
(Take $g=10\, m / s ^{2}$ )

Laws of Motion

Solution:

The block will undergo free fall.
$\Rightarrow s =\frac{1}{2} g(0.2)^{2} $
$=5 \times 0.40=0.2 \,m$