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A 2kg block slides on a horizontal floor...

A `2kg` block slides on a horizontal floor with the a speed of `4m//s` it strikes a uncompressed spring , and compresses it till the block is motionless . The kinetic friction force is compresses is `15 N` and spring constant is `10000 N//m` . The spring by

A

`8.5 cm`

B

`5.5 cm`

C

`2.5 cm`

D

`11.0 cm`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) Let the block compess the spring by `x` before stopping kinetic energy of the block `= (P.E of compressed spring ) + work done against friction .
`(1)/(2) xx 2 xx(4)^(2) = (1)/(2) xx 10000 xx x^(2) + 15 xx x `
`10000 x^(2) + 30x - 32 = 0`
`rArr 50000 x^(2) + 15x - 16 = 0`
`x = (-15 +- sqrt((15)^(2) - 4 xx (50000)(-16)))/(2 xx 5000)`
`:. 0.055m = 5.5 cm`.
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Knowledge Check

  • A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes an uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant is 10000 N/m. The spring is compressed by (in cm):

    A
    2.5
    B
    `11.0`
    C
    `8.5`
    D
    `5.5`

  • A block of mass 4 kg slides on a horizontal frictionless surface with a speed of 2m//s . It is brought to rest in compressing a spring in its path. If the force constant of the spring is 400 N//m , by how much the spring will be compressed

    A
    `2 xx 10^(-2)m`
    B
    `0.2 m`
    C
    `20m`
    D
    `200 m`

  • A block of mass m is initially moving to the right on a horizontal frictionless surface at a speed v. It then compresses a spring of spring constant k. At the instant when the kinetic energy of the block is equal to the potential energy of the spring, the spring is compressed a distance of :

    A
    `vsqrt(m //2k)`
    B
    `vsqrt(m //k)`
    C
    `vsqrt(m //4k)`
    D
    `mv^2//4k`

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