a 10 cm long spring is attached to the ceiling

By Allen Kazadi
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A 10-cm-long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm.What is the spring constant k?How long is the spring when a 4.0 kg mass is suspended from it?

Answer:

As the spring is stretched, it exerts an upward restoring force f. At maximum extension, Newton’s second law gives

∑ F = f – mg = 0   ==>   f = (2.0 kg) (9.8 m/s²) = 19.6 N

By Hooke’s law, if k is the spring constant, then

f = kx   ==>   k = f/x = (19.6 N) / (0.15 m) ≈ 130 N/m

A 4.0 kg mass would cause the spring to exert a force of

f = (4.0 kg) (9.8 m/s²) = 39.2 N

which would result in the spring stretching a distance x such that

39.2 N = (130 N/m) x   ==>   x = (39.2 N) / (130 N/m) ≈ 0.30 m ≈ 30 cm

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