Question

A constant force $F=(\hat{i}+3\hat{j}+4\hat{k})N$ acts on a particle and displace it from $(-1m,2m,1m)$ to $(2m,-3m,1m)$. Then the work done by the force is:

• A. $12J$
• B. $-10J$
• C. $-12J$
• D. $15J$

Answer 1

The correct option is C $-12J$

The position vector of two points $(-1,2,1)$ and $(2,-3,1)$

$\overrightarrow{r}=(2+1)\hat{i}+(-3-2)\hat{j}+(1-1)\hat{k}$

$\overrightarrow{r}=(3\hat{i}-5\hat{j}+0\hat{k})m$

$\overrightarrow{F}=(\hat{i}+3\hat{j}+4\hat{k})N$

The work done by the force,

$W=\overrightarrow{F}.\overrightarrow{r}$

$W=(\hat{i}+3\hat{j}+4\hat{k}).(3\hat{i}-5\hat{j}+0\hat{k})$

$W=1\times 3+3\times (-5)+4\times 0$

$W=3-15=-12J$

The correct option is C.