When a 2.80-kg object is hungvertically on a certain light spring described by H

Question

When a 2.80-kg object is hungvertically on a certain light spring described by Hooke's law, thespring stretches 3.06 cm. (a) What is the force constant of thespring?
1 N/m

(b) If the 2.80-kg object is removed,how far will the spring stretch if a 1.40-kg block is hung on it?
2 cm

(c) How much work must an external agent do to stretch the samespring 6.60 cm from its unstretchedposition?
3 J (a) What is the force constant of thespring?
1 N/m

(b) If the 2.80-kg object is removed,how far will the spring stretch if a 1.40-kg block is hung on it?
2 cm

(c) How much work must an external agent do to stretch the samespring 6.60 cm from its unstretchedposition?
3 J

Explanation / Answer

(a) mass m = 2.80 Kg elongation x = 3.06 cm = 0.0306 m we know that F = k*x                           where k = force constant               Somg = k*x                     k = mg/x                         =(2.80*9.8)/0.0306                         = 896.73 N/m (b) m = 1.4 Kg now elongation in the spring x = mg/k                                             = (1.4*9.8)/896.73                                              =0.0153 m (c) elongation x = 6.60 cm = 0.066 m work done w = 1/2 k x2                     = 0.5*896.73*0.066*0.066                     = 1.953 J

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