A steel wire of length 5m is suspended vertically, stretches by 20mm when mass of 60kg is attached to the lower end. Find the elastic potential energy gained by the wire. (take g=10m/s2)
A
0.6J
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B
6J
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C
60J
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D
600J
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Solution
The correct option is B6J Given, L=5m ΔL=20×10−3m (extension in the wire)
Stretching force (F)=mg=60×10 ⇒F=600N
We know, elastic potential energy of stretched wire, U=12×Stretching force×Extension in wire =12×F×ΔL =12×600×20×10−3 ⇒U=6J ∴ Elastic potential energy gained by the wire is 6J.