A 0.55 kg mass hung from a spring stretches it 0.18 m. (a) How much does it stre

Question

A 0.55 kg mass hung from a spring stretches it 0.18 m. (a) How much does it stretch two such identical springs ifthey are hung in parallel? (b) How much does it stretch the two springs if they are hungin series? (c) Calculate the effective force constant of a singlereplacement spring in each case. (a) How much does it stretch two such identical springs ifthey are hung in parallel? (b) How much does it stretch the two springs if they are hungin series? (c) Calculate the effective force constant of a singlereplacement spring in each case.

Explanation / Answer

Given that m = 0.55 kg x = 0.18m the spring constant k = mg / x                                  = 0.55 kg * 9.8m/s2 / 0.18 m                                  = 29.9 N/m a ) in this case the spring constant k = 2k = 59.88 N/m      Therefore                  x = mg / k = 0.55 kg * 9.8m/s2 / 59.88 N/m                                  = 0.09 m b)  in this case the spring constant k = k /2 = 14.95N/m      Therefore                  x = mg / k = 0.55 kg * 9.8m/s2 / 14.95N/m                                  = 0.36 m      Therefore                  x = mg / k = 0.55 kg * 9.8m/s2 / 14.95N/m                                  = 0.36 m

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