Question

The forces with magnitudes 5,3 and 1 unit are acting on a particle in the direction of the vectors $6\hat{i}+2\hat{j}+3\hat{k};3\hat{i}-2\hat{j}+6\hat{k}$ and $2i-3\hat{j}-6\hat{k},$ respectively. They remain constant while the particle is displaced from the point $A(2,-1,-3)$ to $B(5,-1,1)$. The work done is

• (A) 11 units
• (B) 33 units
• (C) 10 units
• (D) 30 units

Answer 1

The Correct Option is B

Explanation:
$\begin{array}{rl}\because {\overrightarrow{\mathbf{F}}}_1& =\frac{5(6\hat{\mathbf{i}}+2\hat{\mathbf{j}}+3\hat{\mathbf{k}})}{7}\\ & {\overrightarrow{\mathbf{F}}}_2=\frac{3(3\hat{\mathbf{i}}-2\hat{\mathbf{j}}+6\hat{\mathbf{k}})}{7}\\ \text{ and }{\overrightarrow{\mathbf{F}}}_3& =\frac{1(2\hat{\mathbf{i}}-3\hat{\mathbf{j}}-6\hat{\mathbf{k}})}{7}\\ \therefore & \overrightarrow{\mathbf{F}}={\overrightarrow{\mathbf{F}}}_1+{\overrightarrow{\mathbf{F}}}_2+{\overrightarrow{\mathbf{F}}}_3\end{array}$$\begin{array}{rl}& +2\hat{\mathbf{i}}-3\hat{\mathbf{j}}-6\hat{\mathbf{k}})\\ =\frac{1}{7}(30\hat{\mathbf{i}}+10\hat{\mathbf{j}}+15\hat{\mathbf{k}}+9\hat{\mathbf{i}}-6\hat{\mathbf{j}}+18\hat{\mathbf{k}}\\ =\frac{1}{7}[41\hat{\mathbf{i}}+\hat{\mathbf{j}}+27\hat{\mathbf{k}}]\end{array}$$\begin{array}{rl}\text{ and }\overrightarrow{\mathbf{A}\mathbf{B}}& =5\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}-2\hat{\mathbf{i}}+\hat{\mathbf{j}}+3\hat{\mathbf{k}}\\ & =3\hat{\mathbf{i}}+4\hat{\mathbf{k}}\end{array}$$\therefore$ Work done $-\frac{1}{7}[41\hat{\mathbf{i}}+\hat{\mathbf{j}}+27\hat{\mathbf{k}}]\cdot [3\hat{\mathbf{i}}+4\hat{\mathbf{k}}]$$-\frac{1}{7}[123+108]=33$ units