Question

The potential energy (in SI units) of a particle of mass $2\text{kg}$ in a conservative field is $U=6x-8y$. If the initial velocity of the particle 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 $15\text{m}$

$\overrightarrow{F}=-\frac{\partial U}{\partial x}\hat{i}-\frac{\partial U}{\partial y}\hat{j}=-\left[6\hat{i}\right]+\left[8\right]\hat{j}=-6\hat{i}+8\hat{j}$
$\therefore \overrightarrow{a}=-3\hat{i}+4\hat{j}{\text{m/s}}^2$ has same direction as that of $\overrightarrow{u}=\frac{-3\hat{i}+4\hat{j}}{2}\text{m/s}$
So, $\left|\overrightarrow{a}\right|=5{\text{m/s}}^2$,$\left|\overrightarrow{u}\right|=\frac{5}{2}\text{m/s}$
Since $\overrightarrow{u}$ and $\overrightarrow{a}$ are in same direction, particle will move along a straight line.
$S=ut+\frac{1}{2}a{t}^2$
Given $t=2s$
$\therefore S=\frac{5}{2}\times 2+\frac{1}{2}\times 5\times 2^2=5+10=15\text{m}$