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Fall 2008 -- Purdue University -- West Lafayette, IN
Blog Post LABELS
- Blog Challenge Questions (9)
- Chapter 2 (2)
- Chapter 3 (19)
- Chapter 5 (11)
- Chapter 6 (23)
- Chapter 7 (7)
- Chapter 8 (9)
- Exam 1 (5)
- Exam 2 (4)
- Exam 3 (17)
- Exams (2)
- Final Exam (7)
- General (6)
- Gravity (2)
- Homework (2)
- in class (1)
- Lectures (1)
- Quizzes (2)
- Spring 2008 Archive (87)
- Spring-Mass Simulator (1)
- Student-Generated Solutions (1)
- Summer 2008 Archive (72)
10 comments:
I'm not sure how we would be able to do this problem without using the basic potential energy equations that we learned back in high school physics. I solved it with the basic
potential energy = mgh
as my starting point, assumed the system does not lose any energy, and that the rest of the energy is
1/2 k x^2 + 1/2 m v^2
Since we really didn't go over these equations in class, is this right?
Yea I was thinking the same thing, but I'm not sure if its right or not.
--Mark Urbancic
same here guys.
I set whatever the work done by the displacement of spring has to equal to the energy that is exerted by the falling collar.
We can find the work done by the spring by using the equation integral(-kx)dx. The problem gives you values for the k and the displacement of spring. This work is equivalent to the kinetic energy of the falling collar which is 1/2*m*v^2. Then, finally you can solve for v by equating two equations.
I'm not 100% sure if my solution correct, so I accept any suggestion to my work.
I wanted to point out a silly mistake you could make. Remember to convert length to feet or the accleration constant to inches/sec^2 in order for the solution to be dimensionally homogeneous.
Also I would suggest thinking about this problem going from state 1 to state 2. In my setup, I only have potential work from gravity in state 1 and in state 2 i only have work from the spring and KE (b/c there is a velocity). Solve using T_2 = - (V_2 - V_1).
idk if those first three posts work or not, but the 5th one is what I did and it seems like what we did in class.
Another mistake I almost made was in the second step of the problem when calculating the work done from when the spring isn't compressed at all to when it is compressed two inches, you have to have the work done by gravity and by the spring on one side of the equation.
And thanks to steve-o for the tip, I have to go back and change my answer now.
p.s. last comment made by Adam Watson, I still haven't got a blog account and forgot to write my name...
This seems like a good time for the instructor to jump in to offer an opinion.
I would recommend using the work-energy equation of the form T1 + V1 = T2 + V2 (there are no forces other than gravity and the spring that do work, and you can include those in the potential energy) where:
T1 = 0
T2 = (1/2)*m*v^2
V1 = 0 (if you put the gravitational datum at the initial state)
V2 = -mgh + (1/2)*k*delta^2 ("-" sign on the gravitational potential since block is BELOW datum at state 2)
h = (20/12) ft
delta = (2/12) ft
Also, don't forget that 15 lb = WEIGHT (not mass)
Alternately, you can add in the weight and/or spring forces as work (as recommended by geunho above) rather than in potential. This requires integration but is not too bad. However, put these in either work or potential, but NOT both.
Why isn't the equation: (1/2)*k*delta^2 negative? Isn't the spring also below the datum line?
the datum line only affects V_grav
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