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`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 isA. `200 J`B. `100 J`C. zeroD. `1000 J` |
Answer» Net Net work done in sliding a body up to a height `h` on inclined plane `= `Work done against gravitatioinal force + Work done against frictional force ` implies W = W_(g) + W_(f)` (i) but `W = 300J` `W_(g) + mgh = 2 xx 10 xx 10 = 200 J` Putting in Eq (i) we get `300 = 200 + W_(f)` `W_(f) = 300 - 200 = 100J` |
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