physics

A 0.250 kg block on a vertical spring with a spring constant of 5.00 103 N/m is pushed downward, compressing the spring 0.120 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?

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asked by Anonymous
  1. Fs = 5000 N/m * 0.120 m = 600 N.=Force
    of the spring.

    0.5m*V^2 = P.E. = Fs*d,
    0.125*V^2 = 600*0.120,
    0.125V^2 = 72,
    V^2 = 576,
    V = 24 m/s. = Vo.

    h = (Vf^2-Vo^2)/2g,
    h = (0-(24)^2) / -19.6 = 29.4 m.

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    posted by Henry
  2. I CAN'T GET THE RIGHT ANSWER :( A 0.230 kg block on a vertical spring with spring constant of 4.25 multiplied by 103 N/m is pushed downward, compressing the spring 0.048 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?

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    posted by Angela

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