Answer:
0.0928J
Explanation:
the pulling force of spring F=-kx
where x is the displacement from equilibrium position.
energy stored:
[tex]\int\limits^x_0 {-F} \, dx \\=\int\limits^x_0 {kx} \, dx \\\\=\frac{kx^{2} }{2}[/tex]
*** Its fine if you know nothing about calculus. Just apply the equation
[tex]U=\frac{kx^{2} }{2}[/tex]
where U is the potential energy of the spring***
put x=0.150, [tex]U=\frac{8.25}{2}[/tex]×[tex]0.150^{2}[/tex] = 0.0928J (corr. to 3 sig. fig.)
