A person of mass 60 kg is in a lift. The change in the apparent weight of the person, when the lift moves up with an acceleration of 2ms^2and then down with an acceleration of 2ms^2 is
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Answered by
14
Answer:
Explanation:
during upward motion, r-mg=ma , so:
upward acceleration:- R=M(g+a)
= 60(10+2)
= 720 N
during downward motion, mg-r=ma, so:
downward acceleration:- R=M(g-a)
= 60(10-2)
=60 x 8 = 480 N
Answered by
3
Answer:
Explanation:
during upward motion, r-mg=ma , so:
upward acceleration:- R=M(g+a)
= 60(10+2)
= 720 N
during downward motion, mg-r=ma, so:
downward acceleration:- R=M(g-a)
= 60(10-2)
=60 x 8 = 480 N
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