Question

a man of mass 50 kg is standing in an evelator .if the evelator is moving up with an acceleration g/3 the work done by normal reaction of floor of evelator on man when evelator moves by a distance 12 m is (g = 10 m/s2 )​

Answer 1

Answer:

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Answer 2

Answer:

m×a. = force= N. - 50 g

50×g/3. = N. - 50g

N. = 200g /3

Explanation:

Work = F × dis

= 200g/3. × 12

= 8000 J