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Initially the blocks are at rest with F=...

Initially the blocks are at rest with `F=0`. `F` is gradually increased.

From `F=0` till `F=F_(1)`, no motion
From `F gt F_(1)` till `F=2F_(1)`, motion with relative acceleration `=0`
From ` F gt 2F_(1)`, relative acceleration non-zero
At `F=3F_(1)`, relative acceleration `= 2 ms ^(-2)`
Then,

A

`m=2 Kg`

B

`mu_(1)=0.4`

C

`F_(1)=6N`

D

At `F=4F_(1)`, relative acceleration is `4 ms^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
A, D

System starts accelerating when `F=F_(1)`
`F_(1)=0.1 (m+2)g`…(i)
`F_(1)leF le2F_(1)` no relative acceleration, a is common accceleration when `F=2F_(1)`
`2F_(1)-mu_(1)2g=2a….(ii)2F_(1)-mu_(1)2g=2a..(iii)`
`F=3F_(1)` relative acceleration is `2 m//sec^(2)`
`(3F_(1)-mu_(1)2g)/(2)-a=2`...(iv) solving equations
`F_(1)=4,m=2,a=1,mu=0.3`
`F=4F_(1)` relative acceleration is `(4F_(1)-mu_(1)2g)/(2)-a`
putting values relative acceleration is `4 ms^(-2)`
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Knowledge Check

  • A force F_1 acts on a particle so as to accelerate it from rest to a velocity v. The force F_1 is then replaced by F_2 which decelerates it to rest

    A
    `F_1` must be equal to `F_2`
    B
    `F_1` may be equal to `F_2`
    C
    `F_1` must be unequal to `F_2`
    D
    none of these

  • For r gt 0, f(r ) is the ratio of perimeter to area of a circle of radius r. Then f(1) + f(2) is equal to

    A
    1
    B
    2
    C
    3
    D
    4

  • When forces F_(1) , F_(2) , F_(3) are acting on a particle of mass m such that F_(2) and F_(3) are mutually prependicular, then the particle remains stationary. If the force F_(1) is now rejmoved then the acceleration of the particle is

    A
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    B
    `F_(2)F_(3)//mF_(1)`
    C
    `(F_(2)-F_(3))//m`
    D
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