The force exerted by the floor of an elevator on the foot of a person standing there, is more than his weight , if the elevator is
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a
Going down and slowing down
b
Going up and speeding up
c
Going up and slowing down
d
both a and b
answer is D.
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Detailed Solution
When elevator goes up, then equation of motion is R−mg−ma⇒R=m(g+a) i.e., apparent weight > real weight.When elevator goes down and slowing down means retarding a is replaed by -a, then equation of motion is mg−R=m-a⇒R=m(g+a) i.e., apparent weight > real weight.