When a 2.60-kg object is hung vertically on a certain light spring described by

Question

When a 2.60-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.53 cm. (a) What is the force constant of the spring? (No Response) N/m (b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it? (No Response) cm (c) How much work must an external agent do to stretch the same spring 7.90 cm from its unstretched position?

Explanation / Answer

a) F = k*x ===> m*g = k*x ===> k = m*g/x = 2.6*9.8/2.53 = 10.07 N/cm = 1007 N/m b) m*g = k*x ====> x = 1.3*9.8/10.07 = 1.26 cm = 1.26*10^-2 m c) U = (1/2)*k*x^2 = 0.5*1007*(7.9*10^-2)^2 = 3.142J

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