Kevin S. Huang

2011: Problem 15

Topic: Energy
Concepts: Conservation of energy, Vertical spring

Solution:

The initial energy of the mass-spring system is

E_i=\frac{1}{2}k(x+d)^2

where $x$ is the new equilibrium and $d$ is the subsequent displacement. Note we have the initial position of the mass as $U_g=0$ . When the mass passes through the new equilibrium, the energy is

E_f=\frac{1}{2}kx^2+mgd+\frac{1}{2}mv^2

From energy conservation,

\frac{1}{2}k(x+d)^2=\frac{1}{2}kx^2+mgd+\frac{1}{2}mv^2

kxd+\frac{1}{2}kd^2=mgd+\frac{1}{2}mv^2

Since $kx=mg$ ,

\frac{1}{2}kd^2=\frac{1}{2}mv^2

We could have also gotten this directly using the equivalence of a vertical spring to a horizontal spring with shifted equilibrium. Thus,

k=\frac{mv^2}{d^2}=\frac{(0.100\,\mathrm{kg})(0.75\,\mathrm{m/s})^2}{(0.02\,\mathrm{m})^2}=140\,\mathrm{N/m}

so the answer is D.

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Kevin S. Huang

2011: Problem 15

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2011: Problem 15

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