A) 45
B) 75
C) 20
D) 10
Correct Answer: A
Solution :
| Key Idea: Power is equal to the scalar product of force with velocity. |
| Power of the engine, |
| \[P=\vec{F}\,.\,\vec{v}\] |
| Given, \[\vec{F}=(20\hat{i}-3\hat{j}+5\hat{k})\,N\] |
| \[\vec{v}=(6\hat{i}+20\hat{j}-3\hat{k})\,m/s\] |
| Thus, after substituting for \[\vec{F}\] and \[\vec{v}\] in Eq. (i), it becomes, |
| \[P=(20\hat{i}-3\hat{j}+5\hat{k})\,.\,(6\hat{i}+20\hat{j}-3\hat{k})\] |
| \[=(20\times 6)\,(\hat{i}\,.\,\hat{i})+(-3\times 20)\,(\hat{j}\,.\,\hat{j})\,\] |
| \[+(5\,x-3)\,(\hat{k}.\,\hat{k})\] |
| \[=120-60-15\] |
| \[=45\] |
| Note: In the simplification for power, the dot product of a unit vector with same unit vector give 1. |
| The dot product of a unit vector with its orthogonal gives zero. Thus, |
| \[\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1\] |
| \[\hat{i}\cdot \hat{j}=\hat{i}\cdot \hat{k}=\hat{j}\cdot \hat{k}=0\] |
| So, in above simplification second type of dot products are not shown. |
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