300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Work done against friction is (Take g = 10 m/s2)
- 1000 J
- 200 J
- 100 J
- zero
Answer
(3) 100 J
Solution
Work done against gravity = mgh = 2 10 10 = 200 J
∴ Work done against friction = 300 - 200 = 100 J
Answer: (3)