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In the situation shown in figure all con...

In the situation shown in figure all contact surfaces are smooth. The force constant of the spring is K. Two forces F are applied as shown. The maximum elongation produced in the spring is how many times of `F//K` (initially the spring is relaxed)?

Text Solution

Verified by Experts

The correct Answer is:
`(2)`
Maximum elongation is given by
`x_(max)=(2[F_1m_2+F_2m_1])/(K(m_1+m_2))`
Here `F_1=F_2=F`, `m_1=m` and `m_2=M`
Put the values and solve to get `x_(max)=2F//K`
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Knowledge Check

  • Force constant of a spring (k) is anonymous to

    A
    `(YA)/(L)`
    B
    `(YL)/(A)`
    C
    `(AL)/(Y)`
    D
    `ALY`

  • What will be the force constant of the spring system shown in figure?

    A
    `[(1)/(k_(1)) +(1)/(k_(2))]`
    B
    `[(1)/(2k_(1)) +(1)/(k_(2))]^(-1)`
    C
    `[(1)/(k_(1)) +(1)/(k_(2))]^(-1)`
    D
    `[(1)/(2k_(1)) +(1)/(k_(2))]`

  • Two springs have their force constants in the ratio of 3:4 . Both the springs are stretched by applying equal force F. If elongation in first spring is x then elogation is second spring is

    A
    `3x`
    B
    `4x`
    C
    `(4)/(3)x`
    D
    `(3)/(4)x`