When a 2.10-kg object is hungvertically on a certain light spring described by H

Question

When a 2.10-kg object is hungvertically on a certain light spring described by Hooke's law, thespring stretches 2.84 cm. (a) What is the force constant of thespring?
1 N/m

(b) If the 2.10-kg object is removed,how far will the spring stretch if a 1.05-kg block is hung on it?
2 cm

(c) How much work must an external agent do to stretch the samespring 8.20 cm from its unstretchedposition?
3 J (a) What is the force constant of thespring?
1 N/m

(b) If the 2.10-kg object is removed,how far will the spring stretch if a 1.05-kg block is hung on it?
2 cm

(c) How much work must an external agent do to stretch the samespring 8.20 cm from its unstretchedposition?
3 J

Explanation / Answer

   let   k   =   spring(force)constant          x   =   extensionin length of the spring    a.   k   =   F/ x             =   m* g / x             =   2.10* 9.8 / 2.84 * 10-2             =   7.246* 102   N/m             =   7.246* 102   N/m    b.   x   =   F/ k                =   1.05* 9.8 / 7.246 * 102                =   1.42* 10-2   m                =   1.42   cm                =   1.42* 10-2   m                =   1.42   cm    c.   Workdone   W   =   (1/2) *k * x2                                     =   0.5* 7.246 * 102 * (8.20 * 10-2)2                                     =   2.436   J                                     =   2.436   J

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