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Let the two springs A and B be such thatK_(A)gtK_(B) , On which spring will more work has to be done if they are stretched by the same force ? |
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Answer» SOLUTION :`F=K.x` so `x=(F)/(K)` for same `F,W_(A)=(1)/(2)(F^(2))/(K_(A))` and `W_(B)=(F^(2))/(2K_(B)` `therefore (W_(A))/(W_(b))=(F^(2))/(2K_(B))` `AsK_(A)gtK_(B) so W_(A) lt W_(B)`. |
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