What is the stiffness of the rod if it is lengthened by 1 mm under the action of a force of 1000 N?

F = 1000 N.

x = 1 mm = 1 * 10 ^ -3 m.

k -?

According to Hooke’s law, the elastic force F is determined by the formula: F = – k * x, where the “-” sign means that the force is directed in the opposite direction to the elongation, k is the spring stiffness, x is the absolute elongation of the spring.

k = F / x.

k = 1000 N / 1 * 10 ^ -3 m = 1 * 10 ^ 6 N / m.

Answer: the stiffness of the rod is k = 1 * 10 ^ 6 N / m.



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