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Question

The potential energy of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(xāˆ’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(x4)=5x220x
Force, F=dUdx=(10x20)=10x+20
This 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=010x+20=0x=2
So, minimum potential energy will be Umin=5(2)(24)=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=10y
Comparing with
F=kx, k=10N/m
Now, time period of oscillation, T=2πmk=2π0.110=π5sec. (Option D)

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