1. Tardigrade
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  3. Physics
  4. The elastic potential energy of a stretched spring is given by E=50 x2 where x is the displacement in meter and E is in joule, then the force constant of the spring is

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Q. The elastic potential energy of a stretched spring is given by $E=50 x^{2}$ where $x$ is the displacement in meter and $E$ is in joule, then the force constant of the spring is

Work, Energy and Power

Solution:

$E=\frac{1}{2} K x^{2}$