What mass should be suspended from a dynamometer, the spring rate of which is 40 N / m, so that the elongation is 2.5 cm?
Given: spring stiffness k = 40N / m, spring elongation Δl = 2.5 cm = 0.025 m.
Find: the mass of the cargo m -?
Solution: According to Hooke’s law, the modulus of the external force causing the tension of the spring is F = kΔl.
According to Newton’s second law: F = m * g
We get: m * g = kΔl
Let us express m: m = kΔl / g = 40 * 0.025 / 9.8 = 0.102 kg
Answer: the weight of the cargo is 0.102 kg
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