Posted by rei on Saturday, January 14, 2012 at 9:25pm.
A 0.280 kg block on a vertical spring with a spring constant of 4.12 × 10^
3 N/m is pushed downward, compressing the spring 0.0600 m.
When released, the block leaves the spring and travels upward vertically.
The acceleration of gravity is 9.81 m/s^
2. How high does it rise above the point of release?
Answer in units of m

physics  Energy  rei, Saturday, January 14, 2012 at 9:52pm
Just giving the equation to solve the problem would be nice too!

physics  Energy  drwls, Sunday, January 15, 2012 at 6:17am
The potential energy of the compressed spring at release equals the potential energy gained at the maximum distance above the release point. Let that distnce be X and the initial spring compression be d = 0.06 m
(1/2) k d^2 = m g X
k is the spring constant, which you know is 4120 N/m.
m = 0.28 kg
Solve for X
Answer This Question
Related Questions
 Physics  A 0.230 kg block on a vertical spring with spring constant of 4.25e3 N...
 physics  A 0.250 kg block on a vertical spring with a spring constant of 5.00 ...
 physics  A 0.250 kg block on a vertical spring with a spring constant of 5.00 ...
 Physics  A 0.260 kg block on a vertical spring with a spring constant of 5....
 physics  A 0.180 kg block on a vertical spring with spring constant of 4.00 &#...
 Physics  A 0.250 kg block on a vertical spring with a spring constant of 4.09...
 Physics  A 0.340 kg block on a vertical spring with a spring constant of 4.09...
 Physics  A 0.250 kg block on a vertical spring with a spring constant of 4.09...
 PHYSICS  A 0.150 kg block is placed on a light vertical spring (k = 4.60 103 N/...
 Physics  A 540 g block is released from rest at height h0 above a vertical ...
More Related Questions