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Block `A` in the figure is released from rest when the extension in the spring is `x_(0)(x_(0) lt mg//k)`. The maximum downward displacement of the block is (there is no friction) : A. `(2Mg)/(K)-2x_(0)`B. `(Mg)/(2K)+x_(0)`C. `(2Mg)/(K)-x_(0)`D. `(2Mg)/(K)+x_(0)` |
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Answer» Correct Answer - A `(1)/(2)kx_(0)^(2)+Mgh=(1)/(2) k (x_(0)+h)^(2)+0` `rArr h=(2Mg)/(k)-2x_(0)` Maximum downward displacement `=[(2Mg)/(k)-2x_(0)]` |
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