What is the force constant of the spring in a collision between two blocks?

In summary, the conversation revolved around a physics problem involving two blocks colliding and compressing a spring. The conversation included attempts at solving the problem and asking for help, as well as a request for a summary of the conversation. The final output is a revised equation to solve the problem.
  • #1
PhysicsOPhun
14
0
Hi, I Need to know exactly how to solve this...I just can't take it anymore. The only reason I'm so into it now is because I don't get it. I promised my girlfriend I would help her and now she's annoyed because I get more into it than she does. But anyway, someone..PLEASE tell me what has to be done here. I'm sure its something simple but I have no more brain power to even remotely try.



Two blocks, each with a mass 0.40 kg, can slide without friction on a horizontal surface. Initially, block 1 is in motion with a speed v = 1.2 m/s; block 2 is at rest. When block 1 collides with block 2, a spring bumper on block 1 is compressed. Maximum compression of the spring occurs when the two blocks move with the same speed, v/2 = 0.60 m/s.

If the maximum compression of the spring is 1.9 cm, what is its force constant?




I came up with 111N/m but only because I followed a patern, didn't feel like I knew what I was doing...is that right? 111?

 
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  • #2
Well, how did you get that answer; helping you will be easier if we know your way of solving and possible errors in it.
 
  • #3
It would have been better if you had posted your work.

I Need to know exactly how to solve this...I just can't take it anymore

Need some water ??


anyways, its Energy conservation.

[tex]\frac{mv^2_1}{2}=2 \frac{mv^2_{common}}{2}+\frac{Kx^2}{2}[/tex]

K is force constant, x is compression.
 
  • #4
I set the change of potential energy equal to the change in kinetic energy. Then I took final velocity .6m/s and subtracted initial velocity 1.2m/s and set the product of that to equal force constant times the amount the spring collapsed. Having all those values, I calculated for "k".

...?
 
  • #5
I'm sorry, but I didn't get you properly. Can you post your work here.
 
  • #6
{mv^2_{common} - Would this be the value 2/v = .60m/s ? and in terms of mass, do I just double the mass of each box?

Oh, and Sourbah...putting up my work would not help, it's most definatley wrong. I just basically guessed...not much learning there.
 
  • #7
Use the equation which google spider putted up in post #3, I'm not sure you are using the same equation. If you still don't get, just post again with your difficulty
 
  • #8
I have trouble rearranging the variables to solve for the one I need. Also, I don't understand what mv^2common is...

What will this equation look like with my numbers plugged in. I want to work at backwards, understand it conceptually...
 
  • #9
[tex] \frac{mv^2_{common}}{2}[/tex] is the mass of each body (common in this case) multiplied by the velocity (common at the instant of maximum compression); its the KE at the time of maximum compression.

Just write up the values of those variables in the equation which you know, we'll see what remains and how can it be calculated (You may feel irritated with my replies but remember, once I tell you the solution, the fun will be over)
 
  • #10
I'm not irritated, I've already gone through that stage...I'm just in my "got to have the answer" so I can figure out what had to be done to get it. I don't know what the velocity was at the moment of maximum compression. I don't know how to calculate to get those values.

I have:

(.4kg)(1.2m/s)^2 / 2 = 2(.8kg)(1.2) /2 + K(.019m)/ 2

How does that look to you?
 
  • #12
thank you tiny tim...i remember. Really, if there's something that's troubling me...it's very hard for me to get out of the cycle, like my brain shuts down. If I had the answer, and the other values which I don't know...i could work backwards and make perfect sense of it all...
 
  • #13
PhysicsOPhun said:
I'm not irritated, I've already gone through that stage...I'm just in my "got to have the answer" so I can figure out what had to be done to get it. I don't know what the velocity was at the moment of maximum compression. I don't know how to calculate to get those values.

I have:

(.4kg)(1.2m/s)^2 / 2 = 2(.8kg)(1.2) /2 + K(.019m)/ 2

How does that look to you?

Dude, that should look like this :
(.4kg)(1.2m/s)^2 / 2 = 2(.4kg)(0.6)^2 /2 + K(.019m)^2 / 2
 
  • #14
Thanks to everyone who offered their help today... I've spent a week reading a physics book and have been very into it. I'm just doing it for myself so that I could help my friends and family with it if need be. It's a lot harder than I expected but hopefully with some help myself I'll be able to get on top of it soon.
 

Related to What is the force constant of the spring in a collision between two blocks?

What is the force constant of a spring?

The force constant, also known as the spring constant, is a measure of the stiffness of a spring and is represented by the letter k. It is the amount of force required to stretch or compress a spring by a certain distance. The higher the force constant, the stiffer the spring.

How is the force constant of a spring calculated?

The force constant of a spring can be calculated by dividing the applied force by the displacement of the spring. It can also be determined by dividing the potential energy stored in the spring by the displacement squared.

What factors affect the force constant of a spring?

The force constant of a spring is affected by its material, length, and cross-sectional area. In general, a longer and thinner spring will have a lower force constant, while a shorter and thicker spring will have a higher force constant.

What is the relationship between force constant and spring stiffness?

The force constant and spring stiffness are directly proportional to each other. This means that as the force constant increases, the spring becomes stiffer, and it requires more force to stretch or compress it.

How does changing the force constant affect the behavior of a spring?

Changing the force constant of a spring will change its stiffness, which in turn affects how it responds to applied forces. A higher force constant will result in a stiffer spring that is more resistant to deformation, while a lower force constant will result in a softer spring that is easier to deform.

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