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A 75 kg man stands in a lift. What force does the floor exert on him when the elevator starts moving upward with an acceleration of 2 ms^(-2) Given : g = 0 ms^(-2). |
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Answer» Solution :R - mg = ma, `R=mg+ma=m(g+a)` `=75(10+2)N=900N` `=(900)/(10)KG wt. = 90 kg. wt` |
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