The spring potential energy is 1.6 J
Explanation:
The elastic potential energy of a spring is given by the equation:
[tex]U = \frac{1}{2}kx^2[/tex]
where
U is the potential energy
k is the spring constant
x is the compression/stretching of the spring
For the spring in this problem, we have:
x = 40 cm = 0.40 m is the stretching
k = 20 N/m is the spring constant
Substituting into the equation, we can find the spring potential energy:
[tex]U=\frac{1}{2}(20)(0.40)^2=1.6 J[/tex]
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