Question

A body of weight 60 kg is standing in a lift the lift going upward with a uniform velocity of 4.9 metre per second slows down and stops in two seconds then Apparent weight of the body during the slow down process is.​

Answer 1

Answer :-

W = 588 N

Explaination :-

As the lift is goimg upward with uniform velocity, acceleration exerted by the lift will be zero.

Thus, weight of the body will be unaffected by the lift moving with uniform velocity.

Therefore, Apparent Weight of the Man will be -

W = M × G

W = 60 × 9.8

W = 588 N

Therefore, Apparent Weight of the Man during Slow Down Process will be 588 N.

Thanks ....

Answer 2

Answer:

when body is moving upwards

F net= m*(a+g)

=60*(4.9+9.8)

=882 N

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