Finding spring constant K and compression distance

AI Thread Summary
The discussion focuses on solving a physics problem involving the spring constant (K) and compression distance (x) using conservation of momentum and energy principles. Participants emphasize the importance of distinguishing between different physical quantities, such as the maximum force exerted by the spring and kinetic energy. The Work-Energy Theorem is highlighted as a critical tool for relating work done by friction to changes in kinetic energy. A participant successfully calculates the work done by friction and derives the velocities involved in the problem. The conversation concludes with a clearer understanding of how to approach the problem effectively.
mpittma1
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Homework Statement



https://scontent-a-sjc.xx.fbcdn.net/hphotos-frc3/t1.0-9/10294335_1407511836191702_8633020308624670643_n.jpg

Homework Equations


PE=.5kx^2
F=-kx



The Attempt at a Solution



I honestly just don't know how to approach a problem where you have to find both the spring constant and the compression distance...

Ive tried employing conservation of momentum, I've tried working backwards with M1...

Im pretty sure I need to find an equation that would solve for either K or x and then use that to find the other value.

Am I on the right track?
 
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mpittma1 said:

Homework Statement



https://scontent-a-sjc.xx.fbcdn.net/hphotos-frc3/t1.0-9/10294335_1407511836191702_8633020308624670643_n.jpg

Homework Equations


PE=.5kx^2
F=-kx

The Attempt at a Solution



I honestly just don't know how to approach a problem where you have to find both the spring constant and the compression distance...

I've tried employing conservation of momentum, I've tried working backwards with M1...

I'm pretty sure I need to find an equation that would solve for either K or x and then use that to find the other value.

Am I on the right track?
Conservation of momentum and conservation of energy will each be usefully to solve relevant aspects of this problem.


Can you see how you may find the the velocity that m1 attains after the spring fully expands and before m1 gets to the friction surface?
 
SammyS said:
Conservation of momentum and conservation of energy will each be usefully to solve relevant aspects of this problem.


Can you see how you may find the the velocity that m1 attains after the spring fully expands and before m1 gets to the friction surface?

Ya i see that;

100N = .5M1V12

so V1 = 10m/s

right?
 
Start with conservation of momentum to get v2 in terms of v1.

You can get v1 from conservation of energy.
 
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mpittma1 said:
Ya i see that;

100N = .5M1V12

so V1 = 10m/s

right?
Not what I had in mind.

100 N is the maximum force exerted by the spring. (1/2) mv2 is Kinetic Energy. Those are not equal. They're totally different physical quantities.


Calculate the amount of work done by friction in bringing m1 to rest.
 
SammyS said:
Not what I had in mind.

100 N is the maximum force exerted by the spring. (1/2) mv2 is Kinetic Energy. Those are not equal. They're totally different physical quantities.


Calculate the amount of work done by friction in bringing m1 to rest.

Ya i noticed that after i posted.

i will try that though
 
SammyS said:
Not what I had in mind.

100 N is the maximum force exerted by the spring. (1/2) mv2 is Kinetic Energy. Those are not equal. They're totally different physical quantities.


Calculate the amount of work done by friction in bringing m1 to rest.

wf = ukmgd = (.75)(2)(9.8)(2) = 29.4 J
 
would impulse momentum help here?
 
mpittma1 said:
would impulse momentum help here?

Not really.

Are you familiar with the Work - Energy Theorem ? It tells you the relationship between work done and change in Kinetic Energy.
 
  • #10
SammyS said:
Not really.

Are you familiar with the Work - Energy Theorem ? It tells you the relationship between work done and change in Kinetic Energy.

Ya Wnet=ΔKE

So, set 100N = ΔKE of mass 1 and find the initial velocity produced by the spring?
 
  • #11
disregard that last post its the same mistake as the first time...
 
  • #12
K I am getting close;

.5m1v12=wf

v1 = sqrt(29.4) = 5.422

and to find v2

v2=(m1v1/m2) = 1.36m/s
 
  • #13
ugh finally got it!

thanks for all the help!
 
  • #14
Where did that 29.4 value materialize from?
 
  • #15
NascentOxygen said:
Where did that 29.4 value materialize from?
Its the value of the work done by friction

Wf=μk*m*g*d
 
  • #16
mpittma1 said:
Its the value of the work done by friction

Wf=μk*m*g*d
Ah. I can now see that you worked that out earlier.
 
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