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Question

The potential energy of 1 kg particle moving along the X axis is given by U=(x44x22) J. The total mechanical energy of the particle is 2 J. Then maximum speed of the particle is (in m/s)

A
32
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B
12
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C
2
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D
2
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Solution

The correct option is A 32
U=x44x22 (Given)
F=dUdx=(x3x)
dUdx=0x(x21)=0
x=0, ±1
d2Udx2=3x21
d2Udx2=+ve at x=±1
(Point of minima)
i.e Umin=14 J

Kmax+Umin=2 J
P.E. is minimum when kinetic energy is maximum
Kmax=94 J
i.e 12mV2max=94
Vmax=32 m/s
(m=1 kg (given))

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