Under the action of a force of 25N, the spring lengthened by 2 cm.What is the length of the spring under the action of a force of 50 N.
| December 15, 2020 | education
F ctrl = k * / \ l (delta L)
k = F control / / \ l = 25N / 0.02 m = 1275 n / m – we find the coefficient of spring stiffness
/ \ l = F control / k = 50 H * M / 1275 H = 0.04m = 4 cm – we find the delta L given the force given to us in the condition and the stiffness coefficient we found
Answer: 4 cm
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