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A coil spring is extended by 4.2 cm when it supports a 1.00- N weight. If...

A coil spring is extended by 4.2 cm when it supports a 1.00- N weight. If an identical spring is joined to the end of the first one, what is the extension of the combined spring when it supports the same 1.00- N weight? Assume the spring masses to be negligible.

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Answer #1

Hooke's Law

k = F/x = 1.00N/0.042m = 23.80N/m

1/k = 1/k1 + 1/k2
or
k = k1*k2/(k1 + k2)
= 23.80*23.80/(23.8+23.8) = 11.9 N/m

So x = F/k = 1.00/11.9 = 0.0840m = 8.4 cm

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