Answer:
0.20 J
Explanation:
The elastic potential energy stored by a spring is given by:
[tex]U=\frac{1}{2}k(x-x_0)^2[/tex]
where
k is the spring constant
x is the length of the compressed/stretched spring
x0 is the equilibrium length of the spring
In this problem, we have:
k = 150 N/m
x0 = 8.0 cm = 0.08 cm
We need to calculate the elastic potential energy in two situations:
- when x = 11.0 cm = 0.11 m:
[tex]U=\frac{1}{2}(150 N/m)(0.11 m-0.08 m)^2=0.0675 J[/tex]
- when x = 14.0 cm = 0.14 m:
[tex]U=\frac{1}{2}(150 N/m)(0.14 m-0.08 m)^2=0.27 J[/tex]
So, the change in elastic potential energy between the two situations is
[tex]\Delta U=0.27 J-0.0675 J=0.20 J[/tex]
