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.
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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