Question

problem 8:
a ball of mass ( m = 30 mathrm{~g} ) is latched to a compressed a spring. the compression of
the spring is ( x = 5 mathrm{~cm} ), and the spring constant is ( 100 mathrm{~n} / mathrm{m} ). when the latch is
released and therefore the spring is decompressed, what is the final speed ( (v = ?) )
of the ball if it is released horizontally on a frictionless table? use conservation
of energy to solve.

Explanation:

Step1: Convert units

Convert mass \(m = 30\ g=0.03\ kg\) and compression \(x = 5\ cm = 0.05\ m\).

Step2: Apply conservation of energy

The elastic potential energy of the spring \(E_{p}=\frac{1}{2}kx^{2}\) is converted into kinetic energy \(E_{k}=\frac{1}{2}mv^{2}\).
Since \(E_{p} = E_{k}\), we have \(\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}\).
Cancel out \(\frac{1}{2}\) on both sides: \(kx^{2}=mv^{2}\).
Then solve for \(v\): \(v=\sqrt{\frac{kx^{2}}{m}}\).
Substitute \(k = 100\ N/m\), \(x = 0.05\ m\), and \(m=0.03\ kg\) into the formula:
\(v=\sqrt{\frac{100\times(0.05)^{2}}{0.03}}=\sqrt{\frac{100\times0.0025}{0.03}}=\sqrt{\frac{0.25}{0.03}}\approx\sqrt{8.33}\approx 2.9\ m/s\).

Answer:

\(v\approx2.9\ m/s\)