# Homework Help: Dynamics: Spring Question

1. Feb 29, 2016

### jdawg

1. The problem statement, all variables and given/known data
The collar has a mass of 30 kg and is supported on the rod having a coefficient of kinetic friction of 0.4. The attached spring has an unstretched length of 0.2 m and a stiffness of k =50 N/m. Determine the speed of the collar after the applied force F= 200N causes it to be displaced x=1.5 m from point A. When x=0 the collar is held at rest. Neglect the size of the collar in the calculation.

2. Relevant equations

3. The attempt at a solution
I already did the summation of forces in the y direction and found the normal force and used that to find the frictional force.

What I'm really confused about is how to know what length to use for the spring. If someone could explain to me why they use certain lengths for certain potential energies it would help me a lot! I also don't understand why they subtracted 0.5 from 0.2 to get the movable end?

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Last edited: Feb 29, 2016
2. Feb 29, 2016

### BvU

It says spring length is 0.2 m. That's where the moveable end is (call it $x_0$) when the spring is relaxed.
Whole sequence starts at 0.5 m (point A); when the moveable end of the spring is at A the spring is already stretched by 0.3 m.

So for the work calculation you start at $x_i = 0.5$. Spring force is then already $x_i - x_0$.
And the work calculation ends at $x_f= 2$.

3. Feb 29, 2016

### jdawg

Oh ok! Why do they have their x0=0.3 as negative?

And it looks like they integrated from 0 to 1.5. I guess its the same difference. I feel like what you did makes more sense though. Why did they do 0 to 1.5?

4. Feb 29, 2016

### BvU

0.5 m from the wall is taken as x = 0.
(I overlooked that, too, sorry). But, as you say, same difference as long as you indeed correctly write down the differences...

5. Feb 29, 2016

### jdawg

Cool, thanks!!

6. Mar 1, 2016

### Staff: Mentor

Just a reality check: The spring tension during the process is k(x+0.3), and x varies between 0 and 1.5.