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Q.

The potential energy of a 2kg particle free to move along the x-axis is given by U(x)=(x44−x22)J . The total mechanical energy of the particle is 2J . Then the maximum speed (in m/s ) is

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a

2

b

12

c

2

d

32

answer is D.

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Detailed Solution

Kinetic energy is maximum when potential energy is minimum. dUdx=0 ⇒x3−x=0⇒xx2−1=0⇒x2=1 Given KE+PE=2 at x=1,PE=U(1)=(1/4)−(1/2)=-14KE=2−PE=2+(1/4)=(9/4)We know, KE=12mv2=94⇒v=32m/s

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