When a 2.60-kg object is hungvertically on a certain light spring described by H

Question

When a 2.60-kg object is hungvertically on a certain light spring described by Hooke's law, thespring stretches 2.43 cm. (a) What is the force constant of thespring?
1 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?
2 cm

(c) How much work must an external agent do to stretch the samespring 7.70 cm from its unstretchedposition?
3 J (a) What is the force constant of thespring?
1 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?
2 cm

(c) How much work must an external agent do to stretch the samespring 7.70 cm from its unstretchedposition?
3 J

Explanation / Answer

mass m = 2.6 kg Expansion x = 2.43 cm = 0.0243 m (a). force constant k = mg / x                                = 1048.55 N / m (b). Required expasion = Mg / k                                    = ( 1.3 * 9.8 ) / 1048.55                                     =0.01215 m                                    = 1.215 cm (c). external work W = ( 1/ 2) k X ^ 2                                  = 0.5 * 1048.55 * ( 7.7 * 10 ^ -2 m ) ^ 2                                  = 3.108 J

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