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A spring of force constant `800 Nm^(-1)` is stretched initially by `5 cm`. The work done in stretching from `5 cm` to `15 cm` isA. `12.50 N-m`B. `18.75 N-m`C. `25.00 N-m`D. `6.25 N-m` |
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Answer» Correct Answer - C `W = (1)/(2)K(x_(2)^(2)-x_(1)^(2))` |
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