When the spring is stretched by 0.1 m, an elastic force equal to 2.5 N
| June 27, 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 it is stretched by 0.08 m.
x1 = 0.1 m.
Fupr1 = 2.5 N.
x2 = 0.08 m.
En2 -?
The potential energy of a compressed spring Ep is determined by the formula: Ep = k * x ^ 2/2, where k is the stiffness of the spring, x is the elongation of the spring.
En2 = k * x2 ^ 2/2.
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 compressed spring En2 will take the form: En2 = Fupr1 * x2 ^ 2/2 * x1.
En2 = 2.5 H * (0.08 m) ^ 2/2 * 0.1 m = 0.08 J.
Answer: a compressed spring has a potential energy of En2 = 0.08 J.
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