Question

A body is displaced from $\overrightarrow{r_A}=(2\hat{i}+4\hat{j}-2\hat{k})\text{m}\text{to}\overrightarrow{r_B}=(8\hat{i}+14\hat{j}+6\hat{k})\text{m}$ under the action of a constant force $\overrightarrow{F}=(2\hat{i}+3\hat{j}+6\hat{k})\text{N}$.
If the initial kinetic energy of the body is $15\text{J}$, Find the kinetic energy of the body at the end of displacement.

• A. $180\text{J}$
• B. $105\text{J}$
• C. $30\text{J}$
• D. $90\text{J}$

Answer 1

The correct option is B $105\text{J}$

Total displacement of the body is given by $\overrightarrow{S}$:
$\overrightarrow{S}=\overrightarrow{r_B}-\overrightarrow{r_A}$
$=(8\hat{i}+14\hat{j}+6\hat{k})-(2\hat{i}+4\hat{j}-2\hat{k})$
$\therefore \overrightarrow{S}=(6\hat{i}+10\hat{j}+8\hat{k})\text{m}$
work done by the constant force $\overrightarrow{F}$ in displacing the body from position $A$to $B$:
$W_F=\overrightarrow{F}.\overrightarrow{S}$
Or, $W_F=(2\hat{i}+3\hat{j}+6\hat{k}).(6\hat{i}+10\hat{j}+8\hat{k})$
$\therefore W_F=12+30+48=90\text{J}$
$\Rightarrow$Applying work energy theorem on the body from initial to final position:
$W_{\text{All Forces}}=\Delta KE=(KE{)}_f-(KE{)}_i$
$\therefore W_F=(KE{)}_f-15\text{J}$
$90\text{J}=(KE{)}_f-15\text{J}\because \text{given}(KE{)}_i=15\text{J}$
$\therefore (KE{)}_f=90+15=105\text{J}$
So, kinetic energy at the end of displacement is $105\text{J}$.