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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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Knowledge Check

  • Block A in the figure is released from the rest when the extension in the spring is x_(0) . The maximum downward displacement of the block will be :

    A
    `Mg//2k-x_(0)`
    B
    `Mg//2k+x_(0)`
    C
    `2Mg//k-x_(0)`
    D
    `2Mg//k+x_(0)`

  • Block A of the mass M in the figure is released from rest when the extension in the spring is x_(0) . The maximum downwards displacement of the block is (assume x_(0) lt Mg//k ).

    A
    `2 ((Mg)/(k) - x_(0))`
    B
    `(Mg)/(2k) + x_(0)`
    C
    `(2 Mg)/(k) - x_(0)`
    D
    `(2 Mg)/(k) + x_(0)`

  • In above question, speed of block A, when the extension in spring is (x_(m))/(2) , is:

    A
    `2g sqrt((m)/(k))`
    B
    `2 g sqrt((2m)/(k))`
    C
    `2g sqrt((2m)/(3k))`
    D
    `g sqrt((4m)/(3k))`

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