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A person of mass m stands in a lift moving upward with acceleration a. The apparent weight R registered by the lift floor is
Am(g − a)
Bmg only
CZero
Dm(g + a)
Answer & Solution
Correct answer: D. m(g + a)
1. The forces are the upward reaction R and the downward weight mg.
2. For upward acceleration a, the net upward force is R − mg = ma.
3. So R = mg + ma = m(g + a), greater than the actual weight.
4. Rule out m(g − a): that applies to a lift accelerating downward.
_Source: Samacheer Kalvi (TN SCERT) Class 10 Science, Unit 1 "Laws of Motion", p.18_
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