"A 0.200 kg block is placed on a light vertical spring (k = 4.40 103 N/m) and pushed downward, compressing the spring 0.100 m. After the block is released, it leaves the spring and continues to travel upward. What height above the point of release will the block reach if air resistance is negligible?"
How do you do this, and what is your answer? Please break it down into steps. Thanks!
2007-10-09 07:46:09 · 2 answers · asked by justme 4 in Science & Mathematics ➔ Physics
The work done in raising the block to the release point must be accounted for. The spring PE = kx^2/2 is converted to height-change PE = mgΔh, thus Δh = kx^2/(2mg). That's the height change from the compressed-spring position to the maximum height. Height above point of release = Δh - x.
2007-10-10 12:45:46 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
In original state all the energy of the system is potential energy of the string kx^2/2.
In final state spring is relaxed so it has no potential energy,
block's kinetic energy is 0 and its potential energy is mgh.
So mgh = kx^2/2.
h = kx^2/(2mg)
2007-10-09 08:11:39 · answer #2 · answered by Alexey V 5 · 0⤊ 1⤋