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Homework Help: Grade 11 Physics Problem About Work and Energy

  1. Apr 25, 2010 #1
    1. The energy from elastic potential is 1/2 kx^2. K is a spring constant and x is the displacement distance of the spring from its equilibrium position. How far will a spring of spring constant k=25 N/m compress if an obejct of mass 4 kg has an initial velocity of 20 m/s runs into the spring without slowing down or speeding up?

    2. 1/2 kx^2,

    3. Basically, all I could come up with was putting the variables into the equation. To be honest, I don't know where to start solving this question or what equations I should use other than the six kinematic equations. The best I can think of after this is converting the mass into Newtons. Actually, what I wanted to know was how do I start solving this question? Sorry:(
    Last edited: Apr 25, 2010
  2. jcsd
  3. Apr 25, 2010 #2
    Conservation of energy. What is the initial energy?
  4. Apr 25, 2010 #3
    I would take that 20m/s and find the KE(.5mv^2). then that is converted into "spring energy" solve from there.
  5. Apr 25, 2010 #4
    Right. Initially the only energy is the kinetic energy of the object. When it compresses the spring to the maximum distance its final speed will be v = 0 (no kinetic energy). So the only energy at the maximum distance compressed is the potential energy in the spring.
  6. Apr 25, 2010 #5
    hey, just wondering is the answer 17.9m. also the question is not written well. when you say that it doesn't slow down i assumed that it meant that the final velocity is also 20m/s(y/n) if it doesn't slow down at all the spring will be destroyed and carried away with the projectile.
  7. Apr 25, 2010 #6
    It says the INITIAL VELOCITY = 20m/s and it doesn't speed up or slow down. All that means is that it stays at v = 20m/s until it hits the spring. No friction is slowing it down and no other force is speeding it up. Once it hits the spring it does start slowing down as the kinetic energy is converted into potential energy of the spring.
  8. Apr 25, 2010 #7
    yep thats what i thought, do you have an answer for the question?
  9. Apr 25, 2010 #8
    That is not the answer I got. Show your work.
  10. Apr 25, 2010 #9
    Sorry, I don't know if the answer is 17.9m; I wasn't given an answer key. I put the numbers in and got 800 J. I was wondering if the equation Eg=mgh would be any help to figuring out the distance?
  11. Apr 25, 2010 #10
    Your still not showing your work and thus I cannot help you. Why would gravitational potential energy have to do with this system when it is all on a flat table (presumably)? The height does not change in the problem.
  12. Apr 25, 2010 #11
    yeah mgh is static(no movement in the j direction).... 1/2mv^2=1/2kx^2 mv^2=kx^2
    4*20^2=25x^2 x=8m not sure how i got 17 or 19 or whatever..
  13. Apr 25, 2010 #12
    I got 8m as well.
  14. Apr 25, 2010 #13
    I thought there was some kind of relation between height and distance. Right now, all I have is Us=1/2 mv^2, which would be 1/2 (25)(x)^2. Since I'm trying to find x, I'd have to find Us first. But I also have a kinetic energy of 800 J. So I'm guessing I'll have to find an equation that would find Us that includes the kinetic energy? Am I on the right track?
  15. Apr 25, 2010 #14
    No 8m is the answer. Height and distance of what? Do you understand what is happening in this problem? All of the kinetic energy is transferred into elastic potential energy of the spring. At that point the spring is the most compressed and that compression distance is x = 8m. what is Us?
  16. Apr 25, 2010 #15
    Sorry, I didn't refresh the page when I submitted the reply. I think I get it now. But, I was wondering, if the equation is 1/2mv^2, why is the half missing? (Oh, I was trying to look up the notation for potential energy and I got U and s as the sub for the spring, but I'm guessing I looked up the wrong thing)
  17. Apr 25, 2010 #16
    When I say the missing 1/2, I mean when you put then numbers in to solve.
  18. Apr 25, 2010 #17
    [tex] E_i = K_i + Ui = K_i + U_gravity + U_spring = (1/2)M(V_i)^2 + Mgh + 0 [/tex]
    [tex] E_f = K_f + U_f = K_f + U_gravity + U_spring = (1/2)M(V_f)^2 + Mgh + (1/2)kx^2 [/tex]

    [tex]E_i = E_f [/tex]and [tex] V_f = 0 [/tex]

    [tex] (1/2)M(V_i)^2 + Mgh = (1/2)kx^2 + Mgh [/tex]

    Mgh's cancel out. We could have ignored gravitational potential energy from the begining since we know it did not change during the problem.

    [tex] (1/2)M(V_i)^2 = (1/2)kx^2 [/tex]

    Solve for x.
  19. Apr 25, 2010 #18
    do the math there is a half on both sides of the equation so you can cancel them, if you prefer leave them it will give you the same answer. hope i was helpful.
  20. Apr 25, 2010 #19
    Thank you!
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