Under the action of a force of 200 N, the spring was lengthened by 20 cm. What is the stiffness of the spring? How long will this spring lengthen if you act on it with a force of 300 N?
F1 = 200 N.
x1 = 20 cm = 0.2 m.
F2 = 300 N.
When the spring is stretched, the force F with which it is stretched is equal to the elastic force Fel, which arises in the spring: F = Fel.
According to Hooke’s law, the elastic force Fcont, which occurs during deformation, is directly proportional to the elongation of the spring: Fcont = k * x, where k is the coefficient of proportionality, which is called the stiffness of the spring.
F1 = k * x1.
k = F1 / x1.
k = 200 N / 0.2 m = 1000 N / m.
F2 = k * x2.
x2 = F2 / k.
x2 = 300 N / 1000 N / m = 0.3 m.
Answer: the stiffness of the spring is k = 1000 N / m, the spring will stretch by x2 = 0.3 m.
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