NEETPhysicsWork Energy and PowerMCQ+4 / −1
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)
- A1000 J
- B200 J
- C100 J
- Dzero
View written solutionFree
Correct answer: C
Loss in potential energy = mgh
= 2 × 10 × 10 = 200 J.
Gain in kinetic energy = work done = 300 J
Work done against friction = 300 – 200 = 100 J
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