A) \[\frac{149}{7}\]
B) \[\frac{148}{6}\]
C) \[\frac{148}{7}\]
D) None of these
Correct Answer: C
Solution :
Let the forces be \[{{\vec{F}}_{1}}=\frac{5(6\,\hat{i}+2\hat{j}+3\hat{k})}{\sqrt{{{(6)}^{2}}+{{(2)}^{2}}+{{(3)}^{2}}}}=\frac{5}{7}(6\hat{i}+2\hat{j}+3\hat{k})\] and \[{{\vec{F}}_{2}}=\frac{3(3\hat{i}-2\hat{j}+6\hat{k})}{\sqrt{{{(3)}^{2}}+{{(-2)}^{2}}+{{(6)}^{2}}}}\] \[\frac{3}{7}(3\hat{i}-2\hat{j}+6\hat{k})\] \[\therefore \] Total force, \[\vec{F}={{\vec{F}}_{1}}+{{\vec{F}}_{2}}\] \[=\frac{1}{7}(39\hat{i}+4\hat{j}+33\hat{k})\] and let the points be \[\vec{A}=2\hat{i}+2\hat{j}-\hat{k}\] and \[\vec{B}=4\hat{i}+3\hat{j}+\hat{k}\] \[\therefore \] Displacement, \[\vec{d}=\overrightarrow{AB}=2\hat{i}+\hat{j}+2\hat{k}\] \[\therefore \]Work done \[=\vec{d}.\vec{F}\] \[=(2\hat{i}+\hat{j}+2\hat{k}).\frac{1}{7}(39\hat{i}+4\hat{j}+33\hat{k})\] \[=\frac{1}{7}(78+4+66)=\frac{148}{7}unit\]
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