Posted by Anonymous on Sunday, January 15, 2012 at 7:51am.
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?

physics  Henry, Sunday, January 15, 2012 at 4:53pm
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^2Vo^2)/2g,
h = (0(24)^2) / 19.6 = 29.4 m.

physics  Angela, Sunday, October 14, 2012 at 12:26pm
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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