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NEET PHYSICSEasy

A particle moves with a velocity (5i^−3j^+6k^) ms−1(5\hat i-3\hat j+6\hat k)\text{ ms}^{-1} horizontally under the action of a constant force (10i^+10j^+20k^) N(10\hat i+10\hat j+20\hat k)\text{ N}. The instantaneous power supplied to the particle is:

A

200 W

B

zero

C

100 W

D

140 W

Step-by-Step Solution

  1. Identify the Formula: The instantaneous power (PP) is defined as the dot product of the force vector (F\mathbf{F}) and the velocity vector (v\mathbf{v}). P=F⋅vP = \mathbf{F} \cdot \mathbf{v} [Class 11 Physics, Ch 6, Sec 5.10, Eq 5.21]
  2. Substitute Values: F=10i^+10j^+20k^\mathbf{F} = 10\hat i + 10\hat j + 20\hat k v=5i^−3j^+6k^\mathbf{v} = 5\hat i - 3\hat j + 6\hat k
  3. Calculate Dot Product: P=(10)(5)+(10)(−3)+(20)(6)P = (10)(5) + (10)(-3) + (20)(6) P=50−30+120P = 50 - 30 + 120 P=140 WP = 140\text{ W}
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