Question

Potential energy of a body of mass 1 kg, free to move along X-axis, is given by U_(x) = \((\frac{x^4}{4} - \frac{x^2}{2})\)J. Net mechanical energy of the body be 2J, then what will be maximum speed in ms^-1?

Answer 1

Given potential energy of the body of mass m = 1 kg

\(U_{(x)} = (\frac{x^4}{4} - \frac{x^2}{2})\ J\)

The velocity of the body will be maximum when its potential energy will be minimum. i.e.,

\(\frac{dU}{dx} = 0\ and\ \frac{d^2U}{dx^2}\) = Positive

(iii) When x = 1, then \(\frac{d^2U}{dx^2}\) = 3(1)² - 1 = 3 - 1 = 2

Thus it is clear that the potential energy (U) will be minimum when x = -1 or x = 1