Question

A force of $(10\hat{i}-3\hat{j}+6\hat{k})\text{N}$ acts on a body of $5\text{kg}$ and displaces it from $A(6\hat{i}-5\hat{j}+3\hat{k})\text{m}$ to $B(10\hat{i}-2\hat{j}+7\hat{k})\text{m}$. Work done by the force is equal to

• A. zero
• B. $55\text{J}$
• C. $100\text{J}$
• D. $221\text{J}$

Answer 1

The correct option is B $55\text{J}$

Given constant force $\overrightarrow{F}=(10\hat{i}-3\hat{j}+6\hat{k})\text{N}$

Initial position$\left(A\right)$ $\overrightarrow{r_1}=(6\hat{i}-5\hat{j}+3\hat{k})\text{m}$,

Final position$\left(B\right)$ $\overrightarrow{r_2}=(10\hat{i}-2\hat{j}+7\hat{k})\text{m}$

To find the displacement $\overrightarrow{s}$:
$\overrightarrow{s}=\overrightarrow{r_2}-\overrightarrow{r_1}=$ change in position
$\overrightarrow{s}=(10\hat{i}-2\hat{j}+7\hat{k})-(6\hat{i}-5\hat{j}+3\hat{k})$
$\overrightarrow{s}=(4\hat{i}+3\hat{j}+4\hat{k})\text{m}$

Hence the work done , $W=\overrightarrow{F}.\overrightarrow{S}$
$W=(10\hat{i}-3\hat{j}+6\hat{k}).(4\hat{i}+3\hat{j}+4\hat{k})$

$=40-9+24=55\text{J}$

Hence option B is the correct answer