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A heavy-duty stapling gun uses a 0.162-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 32657 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 3.0 10-2m from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.3 10-2m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

  • Physics -

    use this equation
    v = x square root(k/m)
    and the first x

    x=3.0 x 10-2 m
    k=32657 N/m
    m=0.162 kg

    v = x square root(k/m)
    = (0.030m) x ((square root of)32657/0.162
    =(0.030m)(square root of 201586.4198)
    =13.4695129 m/s
    = 13 m/s

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