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Physics Question on work, energy and power

A particle moves with a velocity (5i−3j+6k)ms−1(5i -3j + 6k) ms^{-1} undre the influence of a constant force F=(10i+10j+20K)N.F=(10 i +10 j +20 K)N . The instantaneous power applied to the particle is

A

200 Js−1200\,Js^{-1}

B

40 Js−140\,Js^{-1}

C

140 Js−1140\,Js^{-1}

D

170 Js−1170\,Js^{-1}

Answer

140 Js−1140\,Js^{-1}

Explanation

Solution

Velocity v = (5i - 3j + 6k) ms−1ms^{-1} and force F = (10i + 10j + 20k ) N hence Power P = F. v =(10i+10j+20k)(5i−3j+6k)= (10i + 10j + 20k) (5i -3j + 6k) = 50-30+120 = 140 W = 140Js−1 Js^{-1}