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Question

Out of a pair of identical springs of spring constants 240N/m, one is compressed by 10cm and the other is stretched by 10cm. The difference in P.E. stored in the two springs is:


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Solution

Step 1: Given data

Spring constant K=240N/m

Compression = Elongation = 10cm

Step 2: Formula used

Potential energy stored in spring is given as

P.E=12kx2[whereP.E=potentialenergy,k=springconstant,x=stringdisplacementinm]

Step 3: Calculating P.E. when compressed

The string is compressed by 10 cm. Therefore x=10cm=10100mAs,1cm=1100m

P.Ec=12kx2P.Ec=12×240×101002P.Ec=12010×10P.Ec=1.2J

Step 4: Calculating P.E. when stretched

The string is stretched by 10 cm. Therefore x=10cm=10100mAs,1cm=1100m

P.Es=12kx2P.Es=12×240×101002P.Es=12010×10P.Es=1.2J

Step 5: Calculating the difference in P.E.

Difference in potential energy is as follows

P.E=P.Ec-P.EsP.E=1.2-1.2P.E=0

Hence, the difference in P.E. stored in the two springs is 0.


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