Question: A horizontal spring with stiffness 0.5 N/m has a relaxed length of 15 cm. A mass of 20 g is attached and you stretch the spring to a total length of 25 cm. The mass is then released from rest and moves with little friction. What is the speed of the mass at the moment when the spring returns to its relaxed length of 15 cm?

The magnitude of momentum is equal to the mass times the speed, and the distinct components of momentum along the x, y, and z axes are preserved.

The kinetic energy is given by$\frac{1}{2}m{v}^2$ where mis mass of the object and vis the velocity of the object.

The kinetic energy of spring is given by$\frac{1}{2}k{x}^2$ where k is spring constant and xis the distance from the equilibrium point.

• A horizontal spring with stiffness 0.5 N/m has a relaxed length of 15 cm,
• A mass of 20 g and the spring to a total length of 25 cm.
• To find the speed of the mass at the moment when the spring returns to its relaxed length of 15 cm.

Assume that the system is free of non-conservative forces, therefore energy is conserved between the starting and end states.

Write equation of conservation of energy and use the formula of energy.

$$\begin{gathered} E_1=E_f \\ \frac{1}{2}{k}_s{s}^2=\frac{1}{2}m{v}^2 \end{gathered}$$

Solve the equation for v.

$v=s\sqrt{\frac{k_s}{m}}$

Substitute 0.25-0.15 for s, 0.5for kand 0.020 for minto the obtained equation.

$$\begin{gathered} v=(0.25-0.15)\sqrt{\frac{0.5}{0.020}} \\ =0.5m{s}^{-1} \end{gathered}$$

Therefore, the speed of the mass when the spring returns to its relaxed length is $0.5{\mathrm{ms}}^{-1}$.