JEE Main & Advanced
Sample Paper
JEE Main - Mock Test - 43
question_answer
The potential energy of a 1 kg particle free to move along tile x-axis is given by \[V(x)=\left( \frac{{{x}^{4}}}{4}-\frac{{{x}^{2}}}{2} \right)J.\] The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is