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

`2.0\:\mathrm{J}`

Step-by-step explanation:

The elastic potential energy of a spring is given by:

`U_s=(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=(1)/(2)\cdot 100\cdot 0.2^2,\\U_s=\boxed{2.0\:\mathrm{J}}`