300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. The work done against friction is \[\left( \text{Take g}=\text{ 1}0\text{ m}/{{\text{s}}^{\text{2}}} \right)\]
A) zero
B) 100 J
C) 200 J
D) 300 J
Correct Answer:
B
Solution :
Work done against friction is equal to W = W - U where U is potential energy, W the work done in sliding the block up the inclined plane. U = mgh \[=2\times 10\times 10=200J\] W = 300 J W = 300 - 200 = 100 J