A spring of spring constant 5×103N/m is stretched initially by 5cm from the unstretched position. The work required to further stretch the spring by another 5cm is

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6.25 N-m

12.50 N-m

18.75 N-m

25.00 N-m

answer is C.

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Detailed Solution

work done = change in PE⇒W=Uf−UiW=12K(Xf2−Xi2), here k is force constantx=distance from unstretched positioninitial stretch is 5 cm=0.05meter, final stretch is 10cm=0.10metersubstitute above values , W=12×5×103[0.12−0.052]=18.75J

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