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

Answer:

0.0928J

Explanation:

the pulling force of spring F=-kx

where x is the displacement from equilibrium position.

energy stored:

$\int\limits^x_0 {-F} \, dx \\=\int\limits^x_0 {kx} \, dx \\\\=\frac{kx^{2} }{2}$

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

    $U=\frac{kx^{2} }{2}$

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

  put x=0.150, $U=\frac{8.25}{2}$×$0.150^{2}$ = 0.0928J (corr. to 3 sig. fig.)