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When a 2.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.88 cm. (a) What is the force constant of the spring?

 Dec 5, 2018
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F= -kx    where k = spring constant  x = stretch distance

F=ma = 2 kg x 9.8 m/sec^2 = 19.6 N

 

19.6 = - k (.0288 m)       k = - 680 N/m

 Dec 5, 2018

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