Question

A 0.1 kg mass is connected to a spring with a spring constant of 100 N/m. The spring is
stretched a displacement of 0.2 meters and released. What is the total energy in this system?*
(1 Point)
2.0)
4.0J
10.0)
20.03

Answer 1

Answer:

$2.0\:\mathrm{J}$

Explanation:

The elastic potential energy of a spring is given by:

$U_s=\frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement from equilibrium.

Solving for $U_s$, we get:

$U_s=\frac{1}{2}\cdot 100\cdot 0.2^2,\\U_s=\boxed{2.0\:\mathrm{J}}$