When the spring was lengthened by 5 cm, a force of 60N was expended. Find the stiffness of the spring.

In one of his books, Hooke wrote: “What is stretching, such is strength.” This statement is expressed in the formula

F = k * (L2 – L1),


F is the force applied to the spring;

k is the stiffness of the spring;

(L2 – L1) – the difference between the lengths of the spring before and after the application of the force.

From the formula we find k

k = F / (L2 – L1);

L2 – L1 = 5 cm = 5/100 = 0.05 m;

k = 60 / 0.05 = 1200 N / m

Answer: the spring rate is 1200 N / m.

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