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Question: An engine exerts a force \(\overset{\rightarrow}{F} = (20\widehat{i} - 3\widehat{j} + 5\widehat{k})N...

An engine exerts a force F→=(20i^−3j^+5k^)N\overset{\rightarrow}{F} = (20\widehat{i} - 3\widehat{j} + 5\widehat{k})N and moves with velocity v→=(6i^+20j^−3k^)\overset{\rightarrow}{v} = (6\widehat{i} + 20\widehat{j} - 3\widehat{k}) m/s. The power of the engine (in watt) is –

A

45

B

75

C

20

D

10

Answer

45

Explanation

Solution

Q P=F→.v→=(20i^−3j^+5k^)⋅(6i^+20j^–3k^)P = \overset{\rightarrow}{F}.\overset{\rightarrow}{v} = (20\widehat{i} - 3\widehat{j} + 5\widehat{k}) \cdot (6\widehat{i} + 20\widehat{j}–3\widehat{k}) = 120 – 60 – 15 = 45 W