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Questions  

The potential energy of a 2kg particle free to move along the x-axis is given by U(x)=(x44x22)J . The total mechanical energy of the particle is 2J . Then the maximum speed  (in m/s ) is

a
2
b
12
c
2
d
32

detailed solution

Correct option is D

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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