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A 3.2-kg block is hanging stationary from the end of a vertical spring that is attached to the ceiling. The elastic potential energy of this spring/mass system is 1.7 J. What is the elastic potential energy of the system when the 3.2-kg block is replaced by a 5.3-kg block?

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Added Mon, 11 Jan '16
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At equilibrium

mg =KX

X=mg/K

The elastic Potential energy is

U=(1/2)KX2 =(1/2)*K*(mg/K)2

U=(1/2)(m2g2/K)

so

Potential Energy is proportional to square of the mass

U1/U2=(m1/m2)2

1.7/U2=(3.2/5.3)2

U2=4.66 J

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Added Mon, 11 Jan '16
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