JEE MainPhysicsModerateWork Power & Energy

The potential energy of a particle varies with x according to
the relation $U(x) = x^{2} - 4x$ . The point x=2 is a point of

Show answer and solution

U(x) = x^{2} - 4x

F = 0

\frac{dU(x)}{dx} = 0

2x - 4 = 0\ gives\ \ \ x = 2

\frac{d^{2}U}{{dx}^{2}} = 2 > 0

i.e. U is minimum hence x = 2 is a
point of stable equilibrium.

Source: Grass-G / P-06 Physics paper 14 October.docx

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