1thanx
A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12m/s2. The displacement of the block during the first 0.2 s after the start assuming g = 10 m/s2 is
(a) 4 cm
(b) 20 cm
(c) 24 cm
(d) None of the above
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UP 0 DOWN 0 0 2
2 Answers
21Idea is that acceleration of the lift is 12m/s2 which is greater
than acceleration due to gravity. Hence, when lift starts descending
with this acceleration, block loses contact with the floor and starts
falling under the action of gravity.
Thus, acceleration of the block = acceleration due to gravity = 10 m/s2 (downward)
Hence using s = ut + ½ at2 along downward direction, we
have s = 0 x 0.2 + ½ x 10 x 0.22
i.e. s = 0.2 m = 20 cm.