Answer to Question #122638 in Mechanics | Relativity for Bob Bell

Question

A object with mass 0.688 kg is attached to a horizontal spring with a spring constant of 5.66 N/m. The object slides back and forth on a horizontal frictionless surface. The object is pulled back 5.40 cm and released.
What is the total mechanical energy of the object just as it is released?

How fast is the object traveling when it is a distance 1.78 from equilibrium (presume it is traveling in the positive direction).

Answer 1

The total mechanical energy of the object just as it is released is

$E=\frac{kx^2}{2}=\frac{5.66\cdot0.054^2}{2}=8.25\cdot10^{-3}\text{ J}.$

Assume the mass made 3.68 cm (1.78 cm from the initial equilibrium position). It still has potential energy and some kinetic energy. But initially, all started with elastic potential energy we found above:

$E=E_1+E_K,\\ E=\frac{kx_1^2}{2}+\frac{mv^2}{2},\\\space\\ v=\sqrt{\frac{2E-kx_1^2}{m}}=\\\space\\=\sqrt{\frac{2\cdot8.25\cdot10^{-3}-5.66\cdot0.0178^2}{0.688}}=0.146\text{ m/s}.$