- #1
dnglified1
- 2
- 1
- Homework Statement
- Please see attached file
- Relevant Equations
- Et = Et'
GPE = mgh
SPE = 1/2kx^2
KE = 1/2mv^2
The question asks for a bunch of stuff, but I have everything except part d down.
a) Setting the mass of lemons as m1, I used m1*gh = 1/2mv^2, solving for v of the lemons as v = √2gh, where h is the height at which it is dropped. Then, I used COM and had this equation (not 100% sure if right):
m1*√2gh = (m1 + mp)*vf, and isolated for vf = (m1*√2gh)/(m1 + mp)
b) Setting the equilibrium point at zero, I used kx = mp*g to get x = 0.0054 meters of elongation.
c) Setting Fnet as 0, I used kx = m1*g to get x = m1*g/k
d) This is what I have for the COE equation:
1/2(m1 + mp)vf^2 + mgx = 1/2kx^2
My reasoning for this is that the total energy stored in the spring at max displacement should be equal to the initial kinetic energy of the whole system plus the GPE the spring has due to it not being completely compressed. I solved for x and since x = 2A, A = 0.5x. But I am not getting the expression that is expected. I don't know if the entire thing is wrong or if it is just a math issue.
a) Setting the mass of lemons as m1, I used m1*gh = 1/2mv^2, solving for v of the lemons as v = √2gh, where h is the height at which it is dropped. Then, I used COM and had this equation (not 100% sure if right):
m1*√2gh = (m1 + mp)*vf, and isolated for vf = (m1*√2gh)/(m1 + mp)
b) Setting the equilibrium point at zero, I used kx = mp*g to get x = 0.0054 meters of elongation.
c) Setting Fnet as 0, I used kx = m1*g to get x = m1*g/k
d) This is what I have for the COE equation:
1/2(m1 + mp)vf^2 + mgx = 1/2kx^2
My reasoning for this is that the total energy stored in the spring at max displacement should be equal to the initial kinetic energy of the whole system plus the GPE the spring has due to it not being completely compressed. I solved for x and since x = 2A, A = 0.5x. But I am not getting the expression that is expected. I don't know if the entire thing is wrong or if it is just a math issue.
