Work Energy And Power Question 68

Question: A particle free to move along the x-axis has potential energy given by U(x)=k[1exp(x)2] for x+ , where k is a positive constant of appropriate dimensions. Then [IIT-JEE 1999; UPSEAT 2003]

Options:

A) At point away from the origin, the particle is in unstable equilibrium

B) For any finite non-zero value of x, there is a force directed away from the origin

C) If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin

D) For small displacements from x = 0, the motion is simple harmonic

Show Answer

Answer:

Correct Answer: D

Solution:

Potential energy of the particle U=k(1ex2)

Force on particle F=dUdx=k[ex2×(2x)] F =2kxex2

=2kx[1x2+x42!]

For small displacement F=2kx

therefore F(x)x i.e. motion is simple harmonic motion.



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