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Find the distance the spring compresses

  • Thread starter joemama69
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  • #1
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Homework Statement



a block of mass m slides along ahrizontal table with speed v. At s = 0, it hits a spring with spting constat k and begins to experience a frictin oforce. the coefficient of frictio is u, find the distance the spring compresses when it momentairly comes to rest.

Homework Equations





The Attempt at a Solution



i belive the frictin part is umgs, is this correct

.5mv2 = .5ks2 + ugms

if this is correct, how do i get the s alone
 

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Answers and Replies

  • #2
Doc Al
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i belive the frictin part is umgs, is this correct

.5mv2 = .5ks2 + ugms

if this is correct, how do i get the s alone
It's a quadratic equation. Put it in standard form and solve it.

(That attachment doesn't seem to match the problem you described.)
 
  • #3
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s = -mg + or - (((mg)2 - 4(.5k)(mgh))/k).5
 
  • #4
Doc Al
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s = -mg + or - (((mg)2 - 4(.5k)(mgh))/k).5
How did you get this from the equation you gave in your first post?

I suspect you're solving a different problem.
 
  • #5
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haha ya i had a very similar one sorry about that


s = -umg + or minus (((umg)2 - 4(.5k)(-.5mv2))/k).5
 
  • #6
Doc Al
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s = -umg + or minus (((umg)2 - 4(.5k)(-.5mv2))/k).5
That's almost right. (Redo it more carefully. Each term must have the same units.)

Also, only one of the + or - choices will make sense for this problem.
 
  • #7
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i redid and came up with the same

.5ks2 + umgs - .5mv2 = 0

a = .5k
b = umg
c = -.5mv2

-b - [tex]\sqrt{b^2-4ac}[/tex]/2a

-umg - [tex]\sqrt{()^2 - 4(.5k)(-.5mv^2)}[/tex]/2(.5k)

s = -umg - [tex]\sqrt{()^2 +kmv^2}[/tex]/k
 
  • #8
Doc Al
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-b - [tex]\sqrt{b^2-4ac}[/tex]/2a
Careful with how you do the division by 2a. Everything is divided by 2a, not just the square root term. (Use parentheses to show this.)

And don't forget that it's + or -, not just -.
 
  • #9
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ok

i think it should be minus because the spring is being compressed, do you concure
 
  • #10
Doc Al
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i think it should be minus because the spring is being compressed, do you concure
No...
 
  • #11
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s = -umg + (((umg)2 - 4(.5k)(-.5mv2)).5/k

s = -umg + (((umg)2 + kmv2).5/k
 
  • #12
Doc Al
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s = -umg + (((umg)2 - 4(.5k)(-.5mv2)).5/k

s = -umg + (((umg)2 + kmv2).5/k
You still haven't corrected the "division by 2a" problem from post #8.
 
  • #13
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= (-umg + ((umg)2 + kmv2).5)/k

the answer how ever is Wnc = -m(v0)^2/2 (1 + k/bmg)^-1 after simplifying
 

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