Question

The work done by the force $\overrightarrow{F}=a\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$ in moving the point of application from $(1,1,1)$ to $(2,2,2)$ along a straight line is given to be $5$ units. The value of $a$ is:

• A. $-3$
• B. $3$
• C. $8$
• D. $-8$

Answer 1

Answer: (B)

$3$

Work done by a force $\overrightarrow{F}$ in moving a point thus forming the displacement vector $\overrightarrow{s}$ is given by $W=\overrightarrow{F}.\overrightarrow{s}$

Here, $\overrightarrow{F}=a\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$

Since the point moves from $(1,1,1)$ to $(2,2,2),\overrightarrow{s}=(2-1)\overrightarrow{i}+(2-1)\overrightarrow{j}+(2-1)\overrightarrow{k}$

i.e. $\overrightarrow{s}=\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}$

Work done $=\overrightarrow{F}.\overrightarrow{s}$

$\therefore 5=(a\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}).(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})$

$\therefore 5=a+1+1$

$\therefore a=3$