A spring with k=2000N/m is loaded
with a 50Kg mass. How much energy goesinto stretching the spring?
(remember spring force=-kx)
The amount of energy depends on the extension of the
spring, so we find this first...
x = f/k = (50Kg*9.8m/s/s)/(2000N/m)
= .245m
The mass is just hanging in this question, not oscillating,
so the energy is just the PE,
PE=mgh=50Kg*9.8m/s/s*.245m=120J
The mass from the above question
is tugged downward .2m and released. How fast does it go?
The mass will reach its maximum speed as it crosses
the equilibrium point; assume that's what we are asked to find.
Since the KE maximum corresponds to the PE minimum, we
see that the PE goesinto the KE, and the transfer is 100% at that point...
PE=KE
kx2/2=mv2/2.....
kx2=mv2.....v=sqrt(kx2/m) = sqrt(2000N/m*.2m2/50Kg)=1.3m/s