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Question

# The potential energy of a 4 kg particle free to move along the x− axis is given by U(x)=x33−5x22+6x+3. Total mechanical energy of the particle is 17 J. Then the maximum kinetic energy is

A
10 J
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B
2 J
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C
9.5 J
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D
0.5 J
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Solution

## The correct option is C 9.5 JGiven m=4 kg Potential energy, U(x)=x33−5x22+6x+3 Given: Total mechanical energy E=17 J We know that, E=KE+PE=KE+U ∴ For maximum KE,U should be minimum. i.e dUdx=x2−5x+6=0 ⇒(x−3)(x−2)=0 ∴x=3,2 d2Udx2=2x−5 At x=3,d2Udx2=1>0 (minimum) ∴ Minimum potential energy at x=3 U(x=3)=333−5×92+18+3 Umin=7.5 J ∴(K.E)max=E−Umin =17−7.5=9.5 J

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