Question

The potential energy of a body of mass $2\text{kg}$ in a conservative field is $U=(6x-8y)\text{J}$. If the initial velocity of body is $\overrightarrow{u}=(-1.5\hat{i}+2\hat{j})\text{m/s}$, then the total distance travelled by the particle in first two seconds is

• A. $10\text{m}$
• B. $12\text{m}$
• C. $15\text{m}$
• D. $18\text{m}$

Answer 1

The correct option is C $15\text{m}$

From relation
$\overrightarrow{F}=\frac{-\delta U}{dx}\hat{i}-\frac{\delta U}{\delta y}\hat{j}$
$\Rightarrow \overrightarrow{F}=(-6\hat{i}+8\hat{j})N$
$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{-6\hat{i}+8\hat{j}}{2}=(-3\hat{i}+4\hat{j}){\text{m/s}}^2$
Initial velocity , $\overrightarrow{u}=\frac{-3}{2}\hat{i}+2\hat{j}$
$\Rightarrow \overrightarrow{u}=\frac{-3\hat{i}+4\hat{j}}{2}\text{m/s}$
Hence vector $\overrightarrow{u}\text{and}\overrightarrow{a}$ are in the same direction. Therefore the body will accelerate along a straight line path
$(\because$ angle between $\overrightarrow{a}$ and $\overrightarrow{u}$ is $\theta =0^{\circ })$
$|\overrightarrow{a}|=\sqrt{3^2+4^2}=5{\text{m/s}}^2=\text{constant}$
$|\overrightarrow{u}|=\frac{5}{2}=2.5\text{m/s}$
Applying kinematic equation, the distance travelled by body in $2\text{sec}$ will be:
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow s=(2.5\times 2)+\frac{1}{2}\times 5\times 4$
$\therefore s=15\text{m}$