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In case of an ideal spring, we take the reference point (where potential energy is assumed to be zero) at extension of `x_(0)` instead of taking at natural length. Then potential energy when spring is extended by `x` is `:`A. `(1)/(2)k(x-x_(0))^(2)`B. `(1)/(2)k(x^(2)-x_(0)^(2))`C. `(1)/(2)k(x^(2)-2x x_(0))`D. `(1)/(2)k(2x x_(0)-x^(2))` |
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Answer» Correct Answer - B `DeltaU=U_(x)=U_(x_(0))=(1)/(2)kx^(2)-(1)/(2)kx_(0)^(2)` |
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