Q.6 A man weighing 70 kg stands on a weighing scale which is placed in an elevator. The elevator is
moving up towards its destination floor with a velocity of 1.0 ms–1 . As it approaches the destination
floor it starts slowing down, such that it comes to rest in 2 seconds. Assuming the acceleration due
to gravity, g = 9.8 ms–2 , the reading of the weighing scale just before the elevator comes to rest is
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Elevator weighing scale readings change with acceleration due to Newton’s second law. In this problem, a 70 kg man experiences deceleration while the elevator slows from 1.0 m/s upward to rest in 2 seconds, affecting the scale reading just before stopping.
Problem Analysis
The elevator decelerates upward at a=1.0 m/2 s=0.5 m/s2 downward, since velocity decreases from +1.0 m/s to 0. The scale measures apparent weight N=m(g±a), where upward acceleration increases N and downward decreases it. Here, downward acceleration means N=m(g−a).
Correct Answer Calculation
Apparent weight N=70×(9.8−0.5)=70×9.3=651 N. The reading just before rest is 651 N, as deceleration reduces the normal force below true weight mg=686 N.
Apparent Weight Formula
True weight is mg, but scale reads normal force N from free-body diagram: N−mg=ma, so N=m(g+a) for net upward acceleration (a positive up). Deceleration while going up gives a = -0.5 m/s², yielding lighter reading. Constant velocity (a=0) shows normal mg.
Common Options Explained
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651 N: Correct, matches m(g−0.5) for downward a=0.5 m/s².
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686 N: Wrong; this is mg at rest or constant speed, ignores deceleration.
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721 N: Wrong; assumes upward acceleration of 0.5 m/s², m(g+0.5), opposite to slowing down.
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Zero: Wrong; free fall (a=-g) needed for weightlessness, not this mild deceleration.
Key Physics Insight
During upward travel at constant 1.0 m/s, reading is 686 N. Slowing down mimics downward acceleration, making you feel lighter—like stomach drop in braking elevator. This demonstrates non-inertial frames and effective gravity.
