Conservation of energy and a spring

AI Thread Summary
The discussion revolves around calculating the compression of a spring when a 1.50 kg object is dropped onto it from a height of 1.20 m. The correct approach involves using energy conservation principles, specifically accounting for gravitational potential energy and spring potential energy. Participants clarify that kinetic energy should not be included since the object starts from rest and compresses the spring to a stop. The final calculations lead to a compression of 0.381 m under Earth’s gravity, with adjustments made for scenarios like reduced gravity on the moon and the effect of air resistance. The collaborative effort highlights the importance of correctly defining variables and applying energy conservation in solving physics problems.
Th3Proj3ct
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Homework Statement


A 1.50 kg object is held 1.20 m above a relaxed, massless vertical spring with a force constant of 320 N/m. The object is dropped onto the spring.


Homework Equations


(a) How far does the object compress the spring?
(0.381m)

(b) How far does the object compress the spring if the same experiment is performed on the moon, where g = 1.63 m/s2?
(0.143m)

(c) Repeat part (a), but now assume that a constant air-resistance force of 0.700 N acts on the object during its motion.
(0.371m)


The Attempt at a Solution


I have all the answers already (in parenthesis), but I need help on how to find them. For A I tried
I found velocity with Vf^2=Vi^2+2ad; Vf=Sqrt(2*9.8*1.2); Vf=4.8497
- K+U=K+U
- 0 + .5kx^2 = .5mv^2 + 0
- x = Sqrt(m/k)*v
- x = .332, which isn't the right answer, but I am not sure where I went wrong.
 
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Th3Proj3ct said:

Homework Statement


A 1.50 kg object is held 1.20 m above a relaxed, massless vertical spring with a force constant of 320 N/m. The object is dropped onto the spring.


Homework Equations


(a) How far does the object compress the spring?
(0.381m)

(b) How far does the object compress the spring if the same experiment is performed on the moon, where g = 1.63 m/s2?
(0.143m)

(c) Repeat part (a), but now assume that a constant air-resistance force of 0.700 N acts on the object during its motion.
(0.371m)


The Attempt at a Solution


I have all the answers already (in parenthesis), but I need help on how to find them. For A I tried
I found velocity with Vf^2=Vi^2+2ad; Vf=Sqrt(2*9.8*1.2); Vf=4.8497
- K+U=K+U
- 0 + .5kx^2 = .5mv^2 + 0
- x = Sqrt(m/k)*v
- x = .332, which isn't the right answer, but I am not sure where I went wrong.
looks like you got the final energy on the left side and the initial on the right side, and that -K is shaky, but in any case, you forgot to include the gravitational potential energy term on one side of the equation (which side and with what sign?)
 
I don't think he meant that to be a -K, I think that was just a dash.

You shouldn't have any kinetic energy in the equation, the block falls from rest, and comes to a stop once it reaches the maximum compression.
 
PhanthomJay said:
looks like you got the final energy on the left side and the initial on the right side, and that -K is shaky, but in any case, you forgot to include the gravitational potential energy term on one side of the equation (which side and with what sign?)

Maybe this would have been better:
Kef +Usf +Ugf = Kei + Usi +Ugi
0 + .5kx^2 + 0 = .5mv^2 + 0 + 17.64(mgh = 1.5(9.8)(1.2)=17.64)
But then you get 0.5kx^2=17.64+17.64
x = 4.695

hotcommodity said:
I don't think he meant that to be a -K, I think that was just a dash.

You shouldn't have any kinetic energy in the equation, the block falls from rest, and comes to a stop once it reaches the maximum compression.

Yea it was just a dash, but not using the kinetic, I get
Usf = Ugi
.5kx^2 = mgh
x = sqrt(2mgh/k)
x = 0.332 agian >_<
 
I think you may be defining "h" incorrectly. I'm working on the problem as well and I'm getting a different answer. The initial hight needs to be in terms of the compression distance "x," because that's when the blocks potential energy becomes zero. You should end up with a quadratic equation.
 
Th3Proj3ct said:
Maybe this would have been better:
Kef +Usf +Ugf = Kei + Usi +Ugi
0 + .5kx^2 + 0 = .5mv^2 + 0 + 17.64(mgh = 1.5(9.8)(1.2)=17.64)
But then you get 0.5kx^2=17.64+17.64
x = 4.695
There are always several ways to attack a problem. Stick with one of them. You started by calculating the velocity as the object hit the spring. Fine. If that's your start point, then your mgh term should be mgx, where x is the compressed length of the spring.
 
Thank's all, I combined what you two said and got:
0.5kx^2=17.64+14.7x;
160x^2-14.7x-17.64 which I put into the quadratic formula and got 0.381 as one of the answers. The rest should come pretty easily now. Thanks you two for all the help, I use these forums all the time and you guys really are such a huge help.
 
I know this is an old thread, but why is 14.7 negative in your final quadratic equation?
 
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