When a 4.80 kg object is hungvertically on a certain light spring that obeys Hoo

Question

When a 4.80 kg object is hungvertically on a certain light spring that obeys Hooke's law, thespring stretches 2.50 cm. (a) If the 4.80 kgobject is removed, how far will the spring stretch if a 1.50 kgblock is hung on it?
cm

(b) How much work must an external agent do to stretch the samespring 4.00 cm from its unstretched position?
J (a) If the 4.80 kgobject is removed, how far will the spring stretch if a 1.50 kgblock is hung on it?
cm

(b) How much work must an external agent do to stretch the samespring 4.00 cm from its unstretched position?
J

Explanation / Answer

mass m = 4.8 kg Streached length x = 2.5 cm = 0.025 m We know at equilibrium position , weight = restoringforce                                                         m g = k x from this spring constant k = m g / x                                         = 1881.6 N / m (a). Mass M = 1.5 kg we know   M g = k X from this streached length X = M g / k                                            = 0.0078125 m                                            = 0.78125 cm (b).  work must an external agent do to stretch thesame spring 4.00 cm from its unstretched position is               W = ( 1/ 2) k x ' 2 where x ' = 4 cm = 0.04 m So, W = 1.505 J where x ' = 4 cm = 0.04 m So, W = 1.505 J

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