A sprinf has a potential energy of 84.08 J and a constant of 342.25 N/m. How far it been stretched? Use potential energy elastic
1 answer:
The spring has been stretched 0.701 m
Explanation:
The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as
where
k is the spring constant
x is the elongation of the spring with respect to its equilibrium position
For the spring in this problem, we have:
E = 84.08 J (potential energy)
k = 342.25 N/m (spring constant)
Therefore, its elongation is:
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