Q.

A person of mass m is on the floor of a lift.The lift is moving down with an acceleration ‘a’. Then :a) the net force is acting in downward direction and is equal to mgb) the force mg must be greater than reaction forcec) the man appears to be lighter than his true weight by a factor (a/g)

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a

a, b, c are correct

b

b and c are correct

c

a and c are correct

d

only b is correct

answer is B.

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Detailed Solution

As observed from the ground , net force = (mg-N)=maAs observed from the lift , net force = N+ma -mg=0 ⇒ N = mg-ma < mgApparent weight = N < mg
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A person of mass m is on the floor of a lift.The lift is moving down with an acceleration ‘a’. Then :a) the net force is acting in downward direction and is equal to mgb) the force mg must be greater than reaction forcec) the man appears to be lighter than his true weight by a factor (a/g)