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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

A particle moves with a velocity of (4î−2ĵ+2k̂) m/s under the influence of a constant force F = (10î+3ĵ−2k̂) N. The instantaneous power applied to the particle is W.

Answer: 30

Solution

P = F·v = (10î+3ĵ−2k̂)·(4î−2ĵ+2k̂) = 10×4 + 3×(−2) + (−2)×2 = 40 − 6 − 4 = 30 W.

Question ID VK-154222