Posted by Anonymous on .
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,
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,
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?