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Figure shows a smooth track, a part of w...

Figure shows a smooth track, a part of which is a circle of radius `r`. A block of mass `m` is pushed against a spring of spring constant `k` fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force `mg` when it reaches the point P, where the radius of the track is horizontal.

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Let the initial compression of the spring be x and when the spring is released, the stored compressional energy in the spring is given to the block and it shoots on the track, which ends in a vertical circle of radius `r`. When the mass reaches the point P, the weight of block at this point is in vertical downward direction and the block pushes the track with the only force `mv^2//r`.
At P, the force on track should be equal to the weight of the block, thus
`((mv^2)/(r))=mg` or `v=sqrt(rg)`
Now we apply work-energy theorem from starting point of the spring to the final P for the block.
As initial kinetic energy of the block is zero and at P for the block.
As initial kinetic energy of the block is zero and at P it is `mv^2//2`,
thus we have
`0+1/2kx^2-mgr=1/2mv^2`
`1/2kx^2-mgr=1/2m(rg)`
`kx^2=3mgrimpliesx=sqrt((3mgr)/(k))`
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Knowledge Check

  • In the figure shown a block of masss m is atteched at ends of two spring The other ends of the spring are fixed The mass m is released in the vertical plane when the spring are released The velocity of the block is maximum when

    A
    `k_(1)` is compressed and `k_(2)` is elongated
    B
    `k_(1)` is elongated and `k_(2)` is compressed
    C
    `k_(1)` and `k_(2)` both are compressed
    D
    `k_(1)` and `k_(2)` both are elongated.
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