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Two blocks of equal mass m are connected...

Two blocks of equal mass `m` are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force `F` is applied on the first block pulling away from the other as shown in Fig.

If the extension of the spring is `x_(0)` at time `t`, then the displacement of the first block at this instant is

A

`1/2((Ft^(2))/(2m)+x_(0))`

B

`-1/2((Ft^(2))/(2m)+x_(0))`

C

`1/2((Ft^(2))/(2m)-x_(0))`

D

`(Ft^(2))/(2m)+x_(0)`

Text Solution

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The correct Answer is:
A
a.
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Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force F is applied on one of the blocks pulling it away from the other as shown in figure. (a) Find the displacement of the centre of mass at time t. (b) If the extension of the spring is x_0 at time t, find the displacement of the two blocks at this instant.

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Knowledge Check

  • Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force F is applied on the first block pulling away from the other as shown in Fig. If the extension of the spring is x_(0) at time t , then the displacement of the second block at this instant is

    A
    `(Ft^(2))/(2m)-x_(0)`
    B
    `1/2((Ft^(2))/(2m)+x_(0))`
    C
    `1/2((2F^(2))/m-x_(0))`
    D
    `1/2((Ft^(2))/(2m)-x_(0))`

  • Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. A constant force F is applied on the first block pulling away from the other as shown in Fig. Then the displacement of the centre of mass in at time t is

    A
    `(Ft^(2))/(2m)`
    B
    `(Ft^(2))/(3m)`
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    `(Ft^(2))/(4m)`
    D
    `(Ft^(2))/m`

  • 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)`

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