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Two blocks of masses M = 3 kg and m = 2 ...

Two blocks of masses M = 3 kg and m = 2 kg are in contact on a horizontal table. A constant horizontal force F = 5 N is applied to block M as shown. There is a constant frictional force of 2 N between the table and the block m but no frictional force between the table and the first block M, then acceleration of the two blocks is

A

`0.4 ms^(-2)`

B

`0.6 ms^(-2)`

C

`0.8 ms^(-2)`

D

`1 ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
B
If both blocks move together
`F - f = (m+M) a`
`therefore " "a=(F-f)/(m+M)`
`=(5-2)/(5)=(3)/(5)=0.6 m//s^(2)`
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Knowledge Check

  • Two blocks of masses M = 3 kg and m = 2 kg, are in contact on a horizontal table. A constant horizontal force F = 5 N is applied to block M as shown. There is a constant frictional force of 2 N between the table and the block m but no frictional force between the table and the first block M, then the acceleration of the two blocks is-

    A
    `0.4ms^(-2)`
    B
    `0.6ms^(-2)`
    C
    `0.8 ms^(-2)`
    D
    `1 ms^(-2)`

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