1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

By how much does each spring stretch?

  1. Jan 6, 2007 #1
    1. The problem statement, all variables and given/known data
    Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate to stiffer springs. An object is attached to the lower end of a 14-coil spring that is hanging from the ceiling. The spring stretches by 0.160 m. The spring is then cut into two identical springs of 7 coils each. Each spring is attached between the ceiling and the object. By how much does each spring stretch?

    2. Relevant equations
    F = -kx


    3. The attempt at a solution
    I don't remember an equation that involves the number of coils related to the k value or the distance it stretches. I thought if you cut the spring in half that they'll each stretch to half of the distance. I dont know if thats true. Help is appreciated.
     
    Last edited: Jan 6, 2007
  2. jcsd
  3. Jan 6, 2007 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Welcome to the Forums,

    The big clue is here in the question;
    So, if a spring has 14 coils we can say that;

    [tex]F\propto -\frac{1}{14}x[/tex]

    Therefore, if we have the number of coils in the spring;

    [tex]F\propto - \frac{1}{{\color{red}?}}x[/tex]
     
  4. Jan 6, 2007 #3
    If you produce a spring built from the same material, having the same radius etc. as the referenced spring, than the spring constant is inversely prop. to the number of coils and therefore also inversely prop. to the length of the spring.

    If you half the numbers of coils, you get the doubled spring const. If you attach both new springs the force is also doubled (superposition). If you want to look at this new system of springs like being just one spring, the spring const. k is doubled twice.

    F=-4k x(new)

    Assuming having the same force x(new) is just a 1/4 of the original stretch (40cm)
     
  5. Jan 6, 2007 #4
    so that would be -1/7x, and then do you make them equal to each other because the force is the same? I did that and got an answer of 0.08m for each spring. Is that how you do it?
     
  6. Jan 6, 2007 #5
    yea that makes NW science. Thanks a lot.
     
  7. Jan 6, 2007 #6
    New Problem

    I just have one more question. Theres two parts to it and i got the first part already.

    A 30.0 kg block is resting on a flat horizontal table. On top of this block is resting a 15.0 kg block, to which a horizontal spring is attached. The spring constant of the spring is 315 N/m. The coefficient of kinetic friction between the lower block and the table is 0.555, while the coefficient of static friction between the two blocks is 0.925. A horizontal force F is applied to the lower block as shown. This force is increasing in such a way as to keep the blocks moving at a constant speed.
    At the point where the upper block begins to slip on the lower block determine the following. (a) the amount by which the spring is compressed. (b) the magnitude of the force F.

    For part a) the magnitude by which the spring is compressed is 0.432m.
    For part b) i think there are many factors: the friction of the big block with the table, the friction of the big block with the small block, the force of the spring and how it varies as the force pushes in more. I don't know how to combine all of those into an equation.
     
  8. Jan 6, 2007 #7

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Correct! :smile:
    Not quite. So now you have,

    [tex]F \propto -\frac{1}{x}[/tex]

    However, since the mass is now suspended between two springs each spring shares half the load, i.e. each spring only experiences half the force, therefore;

    [tex]\frac{1}{2}F \propto - \frac{1}{7}x \Rightarrow x \propto \frac{7}{2}x[/tex]

    Do you follow?

    Edit: It is generally not advisable to move onto the next question until you have answered the current one correctly.
     
  9. Jan 6, 2007 #8
    I don't really get that last part with the 7/2x, but i did get the question right using NWsciences way. im not sure if thats the same thing that u were trying to explain or not
     
  10. Jan 6, 2007 #9

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, NWscience has explained the same thing a different way (possibly more lucid). However, note that 0.08 is not the correct answer.
     
  11. Jan 6, 2007 #10
    yea i tried 0.08 the first time but got it wrong, then 0.04 and got it right. Thanks for all your help. I posted up another problem, the only one i have left out of 53 other ones. If you could help on that one too id appreciate it greatly
     
  12. Jan 6, 2007 #11

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For your second question you have calculated the compression in the spring at the point where the block begins to slip. Now, if both blocks are travelling at a constant speed, what can we say about the net forces?
     
  13. Jan 6, 2007 #12
    the net forces are zero because the blocks are not accelerating
     
  14. Jan 6, 2007 #13

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Correc! Now, if we assume both of the blocks move together (as is implied by the question) we can say that the horizontal force applied to the bottom block is equal to the horizontal force applied by the bottom block on the top block, since the applied force is not greater than the maximum frictional force between the two blocks. Do you follow?

    In other words, we can treat the two blocks as a single block. Now, can you write an expression for the forces acting on the block.
     
  15. Jan 6, 2007 #14
    theres the force from the spring which is F = -kx
    theres the force pushing on the 45kg blocks which is just F.which cant be greater than mgu so the blocks dont slip
     
  16. Jan 6, 2007 #15

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, so mathematically;

    [tex]F_{applied} - F_{spring} = 0[/tex]
     
  17. Jan 6, 2007 #16
    so the Force_applied = Force_spring
    and the Force of the spring is -kx
    and k is 315, and is x the answer to a?
     
  18. Jan 6, 2007 #17

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, assuming of course that you answered part (a) correctly.
     
  19. Jan 6, 2007 #18
    i got .432 for the first part, and my k value is 315
    so the force should be 136.08 as im calculating it. i put that in but got it wrong
     
  20. Jan 6, 2007 #19
    and i got part a) correct
     
  21. Jan 6, 2007 #20

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Try rounding to 3sf.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?