With an elongation of 2 cm, the steel spring has a potential energy of elastic deformation of 4 J. How does the potential energy of this spring change when the elongation decreases by 1 cm? Variants of answer: 1) decrease by 1 J 2) decrease by 2 J 3) decrease by 3 J 4) decrease by 4 J
dx1 = 2 centimeters = 0.02 meters – spring elongation;
E1 = 4 Joules – potential energy of the spring at elongation dx1;
dx2 = 1 centimeter = 0.01 meter – spring extension.
It is required to determine dE (Joule) – how much the potential energy of the spring will decrease with lengthening dx2.
Find the coefficient of spring stiffness:
k = 2 * E1 / dx1 ^ 2 = 2 * 4 / 0.02 ^ 2 = 2 * 4 / 0.0004 = 20,000 Newton / meter.
Then the potential energy in the second case will be equal to:
E2 = k * dx2 ^ 2/2 = 20,000 * 0.01 ^ 2/2 = 20,000 * 0.0001 / 2 = 2/2 = 1 Joule.
dE = dE1 – dE2 = 4 – 1 = 3 Joules, that is, it will decrease by 3 Joules.
Answer: the potential energy of the spring will decrease by 3 Joules
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