300 Jof work is done in slide a 2kg bloc...

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

1000 j

200 J

100 J

Zero

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The correct Answer is:

Total work done =Gain in P.E. + Work done against friction
`300=2xx10xx10+WimpliesW=100J`

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