Question

A glider of mass 0.240 kg is on a frictionless, horizontal track, attached to a horizontal spring of force constant 6.00 N/m. Initially the spring (whose other end is fixed) is stretched by 0.100 m and the attached glider is moving at 0.400 m/s in the direction that causes the spring to stretch farther. What is the total mechanical energy (kinetic energy plus elastic potential energy) of the system?

Answer 1

Answer:

$E_M=0.0492J$.

Explanation:

The mechanical energy of the system will be the kinetic energy plus the elastic potential energy: $E_M=K+U_e$.

We know that the equation for the kinetic energy is $K=\frac{mv^2}{2}$, where m is the mass of the object and v its velocity.

We know that the equation for the elastic potential energy is $U_e=\frac{k\Delta x^2}{2}$, where k is the spring constant and $\Delta x$ the compression (or elongation) respect to equilibrium.

So for our values we have:

$E_M=K+U_e=\frac{mv^2}{2}+\frac{k \Delta x^2}{2}=\frac{(0.24kg)(0.4m/s)^2}{2}+\frac{(6N/m)(0.1m)^2}{2}=0.0492J$.