2 By what fraction does the mass of a $m=10\U{g}$, $k=500\U{N/m}$ spring
3 increase when it is compressed by $1\U{cm}$?
9 Spring Su = Spring(pFrom=(0,0), pTo=(4u,0), k=500, L="$m$");
10 Spring Sc = Spring(pFrom=(0,-2u), pTo=(3u,-2u), k=500, L="$m'$");
11 Distance d = Distance(pFrom=(4u,-2u), pTo=(3u,-2u), scale=u, L=rotate(90)*Label("1 cm"));
17 \end{problem} % Based on Q9.11
20 Compression increases the potential energy of the spring by
22 \Delta U = \frac{1}{2} k \Delta x^2
23 = \frac{1}{2} \cdot 500\U{N/m} \cdot \p({0.01\U{m}})^2
26 From Einstein's mass-energy equivalence, increasing the spring's
27 energy must also increase its mass, since mass and energy are two ways
28 of talking about the same stuff.
30 \Delta E &= \Delta m c^2 \\
31 \Delta m &= \frac{\Delta E}{c^2} = \frac{\Delta U}{c^2}
32 = \frac{0.025\U{J}}{(3\E{8}\U{m/s})^2}
33 = 2.78\E{-19}\U{kg} \;.
35 So the fractional mass increase is
37 \frac{\Delta m}{m} = \frac{2.78\E{-19}\U{kg}}{0.010\U{kg}}
38 = \ans{2.78\E{-17}} \;.
41 This mass difference is quite small, which is why it took so long to
42 come up with the $E=mc^2$ idea. Notice, though, that the mass
43 difference is equal to the mass of 16 billion protons (at
44 $1.67\E{-27}\U{kg}$ a pop). Nuclear reactions achieve their high
45 energies through small energy changes for an \emph{enourmous} number
46 of nuclei (on the order of Avogadro's number $N_A =
47 6.022\E{23}\U{particles/mole}$)