A spring (k = 974 N/m) is hanging from the ceiling of an elevator, and a 3.9-kg object is attached to the lowe?

Question

A spring (k = 974 N/m) is hanging from the ceiling of an elevator, and a 3.9-kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at a = 0.41 m/s2?

Answer 1

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The force stretching the spring is: F = m(a + g) = (3.9)(0.41 + 9.8) = 39.8 N

The stretch of the spring: x = F / k = 39.8 / 974 = 0.04 m = 4 cm

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Answer 2

properly there are 2 forces appearing on the mass: the stress tension from the spring and it fairly is weight. subsequently: F=ma which works to T-mg=ma in view that there are 2 forces. T=ma+mg = m(a+g) = 4.7(.sixty six+9.8) = 40 9.2 N = F Now use Hooke's regulation: F= kx x=F/ok = 40 9.2/824 = fifty 9.7 mm or .06 m

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