Q.

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

Moderate

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By Expert Faculty of Sri Chaitanya

The particle is acted upon by a variable force.

The minimum potential energy during motion is -20 J

The speed of the particle is maximum at x = 2m.

The period of oscillation of the particle is (π/5)sec

answer is 1,2,3,4.

(Detailed Solution Below)

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Detailed Solution

F=−dUdx=−5(2x−4) At mean position F=0⇒x=2m Umin=−20J a=−50×2(x−2) ω=10rad/sec T=π/5sec

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