Energy Problem?

Question

A man is standing on a spring. If the spring is compressed 17m, then released...

How high will the man fly?

How fast will he be going when he is released?

If he lands in a hole 10m deep, how fast will he be going when he hits the pillows at the bottom?

Variables:
mass = 80 kg
k = 10 850 N/M
x = 17 m

Answer 1

Using conservation of energy, the energy stored in the spring is equivalent to the potential energy he will gain when he reaches apogee above the coiled spring.

.5*10850*17*17=80*9.8*h
h=(.5*10850*17*17)/(80*9.8)
=2000m
That's really high. Is the spring constant correct at 10850?

The kinetic energy that the man will have minus the loss from gravity in the uncoiling distance when he is ejected.
.5*10850*17*17-80*9.8*17=.5*80*v^2
v=sqrt((.5*10850*17*17-80*9.8*17)/40)
=197m/s

Using an additional gain of 10 m, you can calculate the new velocity for an additional 10m fall.
the total fall is 210 m assuming the top of the coiled spring was even with the ground.

80*9.8*210=.5*80*v^2
v=sqrt((80*9.8*2010)/40)
=198.5 m/s


j

Answer 2

In case you don't remember, the energy stored in a spring is ∫F*ds=∫k*s*s = .5*k^s^2; kinetic energy = .5*m*v^2, and potential energy in gravity field is m*g*h. Kinetic energy at release = .5*k*s^2 - m*g*s. This will equate to the potential energy at his max height. His velocity coming down will be the same magnitude as he had at launch when he passes the launch elevation. The additional velocity he gains falling into the hole comes from the change in potential energy from that point to bottom of the hole converted to kinetic energy.