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Block A in the figure is released from r...

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)`

Text Solution

Verified by Experts

The correct Answer is:
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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