Question

# The potential energy of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(xâˆ’4)J, where x is in meters. It can be concluded that

A
The particle is acted upon by a variable force.
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B
The minimum potential energy during motion is 20 J.
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C
The speed of the particle is maximum at x=2 m.
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D
The period of oscillation of the particle is (π/5) sec.
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Solution

## The correct options are A The particle is acted upon by a variable force. B The minimum potential energy during motion is −20 J. C The period of oscillation of the particle is (π/5) sec. D The speed of the particle is maximum at x=2 m.Potential energy, U=5x(x−4)=5x2−20xForce, F=−dUdx=−(10x−20)=−10x+20This force is dependent on x, hence particle is acted upon by a variable force. (Option A) Now, at equilibrium position potential energy is minimum and kinetic energy is maximum. To determine equilibrium position: F=0−10x+20=0x=2So, minimum potential energy will be Umin=5(2)(2−4)=−20J (Option B)Since kinetic energy is maximum at equilibrium position x=2, the speed of the particle will also be maximum. (Option C)F=−10x+20=−10(x+2)Define a new variable y such that y=x+2. Now, F=−10yComparing with F=−kx, k=10N/mNow, time period of oscillation, T=2π√mk=2π√0.110=π5sec. (Option D)

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