The elastic potential energy of a stretched spring is given by E=50x2 where x is the elongation of the spring in meter and E is in joule, then the force constant of the spring is:
A
50Nm
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B
100Nm−1
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C
100N/m2
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D
100Nm
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Solution
The correct option is A100Nm−1 Elastic potential energy is given by, E=12Kx2GivenE=50x2 Putting in above equation,