Question

A spring of spring constant k=8.25N/m is displaced from equilibrium by a distance of 0.150m. What is the stored energy in the form of spring potential energy? Show your work.

Answer 1

0.0928J

Step-by-step explanation:

the pulling force of spring F=-kx

where x is the displacement from equilibrium position.

energy stored:

`\int\limits^x_0 {-F} \, dx \\=\int\limits^x_0 {kx} \, dx \\\\=(kx^(2) )/(2)`

*** Its fine if you know nothing about calculus. Just apply the equation

`U=(kx^(2) )/(2)`

where U is the potential energy of the spring***

put x=0.150,
`U=(8.25)/(2)`×
`0.150^(2)` = 0.0928J (corr. to 3 sig. fig.)