To deform a spring by 5 cm, a force of 20 N is required. What force must be applied to deform a spring by 12 cm?
| November 27, 2020 | education
F = kx (k-coefficient of stiffness of the spring, x-extension)
k = F / x
5cm = 0.05m
k = 20 / 0.05 = 400 (N / m)
F = kx
12cm = 0.12m
F = 400 * 0.12 = 48 (H)
Answer: 48 N
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