A weight lifter jerks 220 kg vertically ...

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 respectively

`220 J, 330 J`

`3234J, 0 J`

`2334 J, 10 J`

`0 J, 3234 J`

Text Solution

Verified by Experts

The correct Answer is:

`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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