300J of work is done in slinding a 2 kg ...

300J of work is done in slinding a 2 kg block up an inclined plane of height 10m. Taking g =`10m//s^(2)` , work done against friction is

200 J

100 J

Zero

1000 J

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Verified by Experts

The correct Answer is:

By using work energy theorem W`=DeltaKE
(hereDeltaKE=0)implies300-W_(gravity)-W_("friction")=0`
`W_("friction")=300-mgh`
`=300-(2)(10)(10)=100J`

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