A) 11 unit
B) 33 unit
C) 10 unit
D) 30 unit
Correct Answer: B
Solution :
Key Idea: If the force \[\vec{F}\]and displacement \[\vec{d},\]then work done \[=\vec{F}.\vec{d}.\] Since, \[{{\vec{F}}_{1}}=\frac{5(6\hat{i}+2\hat{j}+3\hat{k})}{7}\] \[{{\vec{F}}_{2}}=\frac{3(3\hat{i}-2\hat{j}+6\hat{k})}{7}\] and \[{{\vec{F}}_{3}}=\frac{1(2\hat{i}-3\hat{j}-6\hat{k})}{7}\] \[\therefore \]Resultant force, \[\vec{F}={{\vec{F}}_{1}}+{{\vec{F}}_{2}}+{{\vec{F}}_{3}}\] \[=\frac{1}{7}(30\hat{i}+10\hat{j}+15\hat{k}+9\hat{i}-6\hat{j}+18\hat{k}+2\hat{i}-3\hat{j}-6\hat{k})\] \[=\frac{1}{7}(41\hat{i}+\hat{j}+27\hat{k})\] Now, \[\overrightarrow{AB}=(5i-\hat{j}-\hat{k})-(2\hat{i}-\hat{j}-3\hat{k})\] \[=3\hat{i}+0\hat{j}+4\hat{k}\] \[\therefore \]Work done \[\mathbf{=\vec{F}}\mathbf{.}\overrightarrow{\mathbf{AB}}\] \[=\frac{1}{7}(41\hat{i}+\hat{j}+27\hat{k}).(3\hat{i}+4\hat{k})\] \[=\frac{1}{7}(123+108)=33unit\]You need to login to perform this action.
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