Question

Forces of magnitude $3$ and $4$ units acts along $6\hat{i}+2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+6\hat{k}$ respectively on a particle and displaces it from $(2,2,-1)$ to $(4,3,1)$, then the work done is

• A. $\frac{124}{7}$
• B. $\frac{120}{7}$
• C. $\frac{115}{7}$
• D. None

Answer 1

Answer: (A)

$\frac{124}{7}$

Total force
$\overrightarrow{F}=3\left(\frac{6\hat{i}+2\hat{j}+3\hat{k}}{7}\right)+4\left(\frac{6\hat{i}+2\hat{j}+3\hat{k}}{7}\right)=\frac{1}{7}(30\hat{i}-2\hat{j}+33\hat{k})$
displacement $\overrightarrow{d}=2\hat{i}+\hat{j}+2\hat{k}$
$W=\overrightarrow{F}.\overrightarrow{d}=\frac{124}{7}$