When a 2.20-kg object is hung vertically on a certain light spring described by

Question

When a 2.20-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 3.20 cm. (a) What is the force constant of the spring? 673.75 N/m (b) If the 2.20-kg object is removed, how far will the spring stretch if a 1.10-kg block is hung on it? 00016 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. cm (c) How much work must an external agent do to stretch the same spring 6.50 cm from its unstretched position? J

Explanation / Answer

Answer to the above question is as follows

By Hooks's Law

a) F = k x here the force is the weight of the object so F = m g and we have

m g = k x so that k = m g/x = 2.2kg x 9.81m/s/s / 0.0320m = 674.43N/m

b) if half the weight is put on the spring, it will stretch the spring half as far, or 1.60cm(0.0160m)

x=mg/k = 1.1 x 9.81/674.43 = 0.0160m

c) W = 1/2 k x^2 = 1/2 x 674.43N/m (0.0650m)^2 = 1.424J

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