The potential energy of a particle of mass 1 kg in a conservative field is given as $U = (3 x^{2} y^{2} + 6 x) J,$ where $x$ and $y$ are measured in meter. Initially particle is at (1, 1) & at rest then :

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

Work done by external agent to slowly bring the particle to origin is $9 J$

Work done by external agent to slowly bring the particle to origin is $- 9 J$

If particle is left free it moves in straight line

Initial acceleration of particle is $6 \sqrt{5} m s^{- 2}$

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$F = - \frac{\partial U}{\partial x} \hat{i} - \frac{\partial U}{\partial y} \hat{j}$
$F = - [6 x y^{2} + 6] \hat{i} - (6 x^{2} y) \hat{j}$ at $(1, 1)$

$F = - [12] \hat{i} - 6 \hat{j} = - 12 \hat{i} - 6 \hat{j}$
$a = \frac{F}{M} = - 12 \hat{i} - 6 \hat{j}$

$| a | = \sqrt{12^{2} + 6^{2}} = 6 \sqrt{5} m / s^{2}$
Work done by conservative force $= U_{i} - U_{f}$
$= (3 + 6) - 0$
$= 9 J$
Hence work done by
External agent $= U_{f} - U_{i}$
$= - 9 J$