Under the action of a weight of 200 g, the spring was stretched by 5 cm to determine the spring stiffness.
| December 21, 2020 | education
Given:
m = 200 g = 0.2 kg
X = 5 cm = 0.05 m
k =?
Decision:
F = kx (elastic force)
mg = kx
0.2 * 10 = k * 0.05
2 = 0.05k
k = 2 ÷ 0.05
k = 40
Answer: spring rate is 40 N / m
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