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Q.

The potential energy of a 1 Kg particle free move along the x-axis is given by V(x)=(x44−x22)J . The total mechanical energy of the particle 2J. Then ,the maximum speed (in m/s) is

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a

2

b

3/2

c

2

d

1/2

answer is B.

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Detailed Solution

U=x44-x22 F=dUdx=4x34-2x2=x3-x 0=xx2-1 x=0,±1 d2Udx2=3x2-1 For x=±1, d2Udx2=+iveK.Emax=ET−UminUmin(±1)=−1/4JKEmax=121v2=9/4J ⇒v=32J
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