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A weight lifter jerks `220 kg` vertically through `1.5` metre and holds still at that height for two minutes. The work done by him in lifting and in holding it still are respectivelyA. `220 J, 330 J`B. `3234J, 0 J`C. `2334 J, 10 J`D. `0 J, 3234 J` |
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Answer» Correct Answer - B `W = vec(F).vec(S) = FS cos theta` In lifting the weight `F = mg, theta = 0^(@)`, in holding the weight, `S = 0` |
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