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Spring forces, etc

  • Thread starter alex9050
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  • #1
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Spring forces, etc.... :(

Homework Statement



Doing past exams before a real exam and i completely missed this whole section of lectures. lol

A Block of mass 0.800kg is given an intitial velocity v = 1.2m/s to the left and collides with a light spring of force constant k = 50.0N/m


Homework Equations



If the surface is frictionless, calculate the maximum compression of the spring after the collision.

The Attempt at a Solution



I have absolutely no idea how to begin this questions. any help with these sorts of questions or this one in particular would be great
 

Answers and Replies

  • #2
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Hey alex9050. Welcome to PF.

First off, try think what the block will be doing when the spring is fully compressed. At what speed will it be traveling?
 
  • #3
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http://cnx.org/content/m14102/latest/

scroll to the bottom... theres almost exactly the same problem except with different numbers... read it carefully because i believe they made an error by putting -250 instead of -500 at the bottom... just plug in your numbers and try and understand the work energy laws
 
  • #4
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Sorry, what are you trying to look for in your original question?
 
  • #5
Im not really sure what the question is... but just based on the i would start by finding the kinetic energy of the block, since we know the surface is frictionless, think about where all that energy has to go when it hits the spring.
 
  • #6
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we're trying to calculate the maximum compression of the spring made by the block before the block goes in the other direction

and ed, i do understand the block will be stationary for a split second, but i have no idea on the formula to use to answer the question :$
 
Last edited:
  • #7
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http://cnx.org/content/m14102/latest/

scroll to the bottom... theres almost exactly the same problem except with different numbers... read it carefully because i believe they made an error by putting -250 instead of -500 at the bottom... just plug in your numbers and try and understand the work energy laws
and this is a work done question.
 
  • #8
Dick
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Equate the kinetic energy in the block to the potential energy in the spring at maximum compression, since then the block will have no kinetic energy, and the energy must have gone somewhere.
 
  • #9
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so that means that its 1/2kx^2 = 0 ? then just rearrange and solve ? because the KE = .5mv^2 = 0 because velocity is 0 at the point of maximum compression ?
 
  • #10
but i have no idea on the formula to use to answer the question :$
the equations you will want to use are

Ek= (1/2)(m)(v^2 - u^2) where u is the initial velocity
Eel= (1/2)(constant)(Δx^2) where Δx is how far the spring is compressed.

see if you can figure out where to go from there.
 
  • #11
Dick
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so that means that its 1/2kx^2 = 0 ? then just rearrange and solve ? because the KE = .5mv^2 = 0 because velocity is 0 at the point of maximum compression ?
Nooooo. (1/2)*kx^2 is equal to the INITIAL kinetic energy. At the point where v=0 all of the initial kinetic energy has been transferred to potential energy in the spring.
 
  • #12
so that means that its 1/2kx^2 = 0 ? then just rearrange and solve ? because the KE = .5mv^2 = 0 because velocity is 0 at the point of maximum compression ?
yes the velocity will be 0 at the point of maximum compression. but thats not what you want to use. you want to find the kinetic energy before it compresses the spring. so use the .5mv^2 before it gets to the spring and set things equal. sorry for the last post, i didnt see you had the formulas. but find the kinetic energy before it ever gets to the spring.
 
  • #13
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wait wait, working it out now.
 
  • #14
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i got 0.152 so 0.152M?
 
  • #15
it would be positive because the compression in the same direction as the velocity you use.
 
  • #16
Dick
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  • #17
that is what i got too.
 
  • #18
Dick
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  • #19
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so i ended up as .5*.8*(1.2)^2 = .5*50 *x^2 does that sound right ?

equaling 0.152m
 
  • #20
Dick
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I think we are all agreed that that's right.
 
  • #21
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AWESOME! thanks guys :)
 

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