A) \[\frac{Mg}{2k}+\frac{mg}{k}\]
B) \[\frac{2Mg}{k}+\frac{mg}{2k}\]
C) \[\frac{Mg}{k}+\frac{2mg}{k}\]
D) \[\frac{Mg}{k}+\frac{mg}{k}\]
Correct Answer: D
Solution :
[d] The block will lose contact if the spring compresses more than \[\frac{Mg}{k}\]. In equilibrium position the spring is stretched by \[\frac{mg}{k}\] . If the spring oscillates with amplitude A, it will move up above its equilibrium position by A. It means compression in this extreme win position will be \[A-\frac{mg}{k}\]. For the block to remain on the table, \[A-\frac{Mg}{k}=\frac{mg}{k}\] \[A=\frac{Mg}{k}+\frac{mg}{k}\]
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