Kevin S. Huang

2012: Problem 8

Topic: Energy
Concepts: Conservation of energy, Energy dissipation, Spring-potential energy

Solution:

By conservation of energy, the initial kinetic energy of the block is converted to spring potential energy and energy dissipated by friction. We have

K=U_{\text{spr}}+E_{\text{dis}}

\frac{1}{2}mv^2=\frac{1}{2}kx^2+\mu_kmgx

kx^2+(2\mu_kmg)x-mv^2=0

Solving for $x$ using the quadratic equation,

x=\frac{-2\mu_kmg+\sqrt{(2\mu_kmg)^2-4k(-mv^2)}}{2k}=-\frac{\mu_kmg}{k}+\sqrt{\left(\frac{\mu_kmg}{k}\right)^2+\frac{mv^2}{k}}=0.24\,\mathrm{m}

where we took the positive root since $x>0$ under our sign convention. Thus, the answer is B.

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2012, F=ma Exam

Conservation of energy, Energy dissipation, Fma: Energy, Spring potential energy

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