When the spring is stretched by 0.1 m, an elastic force equal to 2.5 N. arises in it.
| August 30, 2021 | education
When the spring is stretched by 0.1 m, an elastic force equal to 2.5 N. arises in it. Determine the potential energy of this spring when stretched to 0.08m.
x1 = 0.1 m.
Fcont1 = 2.5 N.
x2 = 0.08 m.
Wп -?
The potential energy of a stretched spring Wp is determined by the formula: Wp = k * x2 ^ 2/2, where k is the stiffness of the spring, x2 is the change in the length of the spring during stretching.
We express the stiffness of the spring k from Hooke’s law: Fupr1 = k * x1.
k = Fcont1 / x1.
The formula for determining the potential energy of a stretched spring will take the form: Wp = Fupr1 * x2 ^ 2/2 * x1.
Wp = 2.5 N * (0.08 m) ^ 2/2 * 0.1 m = 0.09 J.
Answer: an extended spring at x2 = 0.08 m has a potential energy Wp = 0.09 J.
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