Question

1. If the spring of a jack-in-the-box is compressed a distance of 8.00 cm from its relaxed length and then released, what is the speed of the toy head when the spring returns to its natural length? Assume the mass of the toy head is 50.0 g, the spring constant is 80.0 N/m and the toy head moves only in the vertical direction. Also disregard the mass of the spring. (Hint: Remember that there are two forms of potential energy in the problem.)

Answer 1

When the spring is compressed, the total energy of the jack-in-the-box system is elastic potential energy stored in the spring:
$E_i= \frac{1}{2} kx^2$
where $k=80.0 N/m$ is the spring's constant and $x=-8 cm=-0.08 m$ is the displacement of the spring with respect its rest position.

When the spring is released and it reaches its relaxed position, its elastic energy becomes zero (because x=0), and so the total energy of the system will be the kinetic energy of the toy's head:
$E_f= \frac{1}{2} mv^2$
where $m=50 g=0.05 kg$ is the mass of the head and v its velocity.

For the law of conservation of energy, $E_i = E_f$. Rewriting both terms, we can find v:
$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$
$v= \sqrt{ \frac{kx^2}{m} }=3.2 m/s$