The elastic potential energy of a stretched spring is given by E = 50x2 . Where x is the displacement in meter and E is in joule, then the force constant of the spring is
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a
50 Nm
b
100 Nm-1
c
100 N/m2
d
100 Nm
answer is B.
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Detailed Solution
U=12Kx2--(1), U=50x2--(2), compare equation (1) and (2) to find K