JEE Main · PhysicsHard

$A particle moves along the x-axis under the influence of a force $F = -kx + \alpha x^3$. What is the potential energy function $U(x)$ if $U(0) = 0$?$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{1}{2}kx^2 - \frac{1}{4}\alpha x^4$

Solution

The potential energy is defined as

U (x) = - \int_{0}^{x} F (x^{'}) d x^{'}

. Integrating

F = - k x + α x^{3}

gives

U (x) = \frac{1}{2} k x^{2} - \frac{1}{4} α x^{4}