How many times the length of the spring with a stiffness of 150N / m increased, if a tensile force of 50N is applied to it. The length of the spring in an undeformed state is 15 cm.
Initial data: k (coefficient of spring stiffness) = 150 N / m; F (the amount of tensile force applied to the spring) = 50 N; l (spring length in an undeformed state) = 15 cm = 0.15 m.
1) Determine the elongation of the spring: Δl = F / k = 50/150 = 0.33 m.
2) Calculate the length of the spring after stretching: l1 = l + Δl = 0.15 + 0.33 = 0.48 m.
3) Increase in the length of the spring: l1 / l = 0.48 / 0.15 = 3.2 times.
Answer: The length of the spring after deformation increased by 3.2 times.
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