Question

A 0.300 kg mass is attached toa 26.6 N/m spring. It is pulled0.120 m and released. How much Potential Energy does it have when it is 0.0600 m from equilibrium?(Unit = J)

Answer 1

0.144 Joules

Explanation

elastic potential energy is given by

$\begin{gathered} PE=\frac{1}{2}kx^2 \\ where\text{ K is the spring constatn} \\ x\text{ is the amount of compresion \lparen}relative\text{ to equilibrium\rparen} \end{gathered}$

so

Step 1

Diagram:

given

$\begin{gathered} m=0.3\text{ kg} \\ k=26.6\frac{N}{m} \\ x=0.120\text{ m} \end{gathered}$

Step 1

a) energy when the spting is compressed

Total elastic energy of the spring = Kinetic energy of the spring + potential energy of the spring

so

$\begin{gathered} E=\frac{1}{2}kx^2=\frac{1}{2}(26.6\text{ N/m\rparen\lparen0.12}^2)=0.19152\text{ J} \\ EP=\frac{1}{2}kx^2=\frac{1}{2}(26.6)(0.6^2)=0.04788 \\ hence \\ Ek=0.19152\text{ J-0.04788} \\ Ek=0.144\text{ Joules} \end{gathered}$

the kinetic energy is given by