When a 2.40-kg object is hung verticallyon a certain light spring described by H

Question

When a 2.40-kg object is hung verticallyon a certain light spring described by Hooke's law, the springstretches 3.19 cm. (a) What is the force constant of thespring?
1 N/m

(b) If the 2.40-kg object is removed, howfar will the spring stretch if a 1.20-kgblock is hung on it?
2 cm

(c) How much work must an external agent do to stretch the samespring 8.70 cm from its unstretchedposition?
3 J (a) What is the force constant of thespring?
1 N/m

(b) If the 2.40-kg object is removed, howfar will the spring stretch if a 1.20-kgblock is hung on it?
2 cm

(c) How much work must an external agent do to stretch the samespring 8.70 cm from its unstretchedposition?
3 J

Explanation / Answer

mass m = 2.4 kg streached length x = 3.19 cm = 0.0319 m force constant K = F / x                           = mg / x                           = 737.3 N / m (b). streached length x ' = Mg / K                                     = 1.2 * 9.8 / ( 737.3 )                                     = 0.01595 m                                     = 1.595 cm (c). work = ( 1/ 2) K x " ^ 2               = 0.5 * 737.3 * ( 8.7 cm ) ^ 2               = 0.5 * 737.3 * ( 0.087 m ) ^ 2               = 2.79 J

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