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Physics spring constant

  1. Nov 23, 2009 #1
    A 1300-kg car starts at rest and rolls down a hill from a height of 10.0-m. It moves across a level surface and collides with a spring-loaded guard rail designed to bring the car safely to a stop. The spring has a spring constant of 2.0E6N/m.
    Neglecting any losses due to friction, and ignoring the rotational kinetic energy of the wheels, find the maximum distance the spring is compressed.

    Calculate the maximum acceleration of the car after contact with the spring (again ignore brakes, friction, etc.). Comment on the possibility of injury to the driver.

    If the spring is compressed by only 0.30-m, find the change in the mechanical energy due to friction.

    *I am really not sure where to start with this quesion can someone please help me?
     
  2. jcsd
  3. Nov 23, 2009 #2
    [tex]E_o = E_f[/tex]
    That will get you the first part
     
  4. Nov 23, 2009 #3
    First you should be asking yourself, "what is going on?". You have an object starting from rest at a certain height, and then it proceeds downward toward a flat surface. Energy is needed to make the object move. Where does this energy come from? Is energy conserved? And finally, What happens to this energy as the object moves down the hill and compresses the spring?

    As for the second part, you are analyzing the part where the car is traveling on the flat surface, and then continues to compress the spring. You can solve this part using F = ma. What forces are acting on the car? We know a spring force is acting on it.. when is this spring force maximum?
     
  5. Nov 23, 2009 #4
    [tex] E_o = E_f [/tex]
    [tex] mgh = \frac {1}{2} k x^2[/tex]
    I don't know why you have mgx in the second part. The car is moving horizontally once it hits the spring.
     
  6. Nov 23, 2009 #5
    oh right...so it would be
    x= squar(127400(2)/2.0E6)=.4 m
     
    Last edited: Nov 23, 2009
  7. Nov 23, 2009 #6
    You messed up on your calculations.
     
  8. Nov 23, 2009 #7
    How do i find the maximum acceleration?
     
  9. Nov 23, 2009 #8
    F = -kx = ma
     
  10. Nov 23, 2009 #9
    Is this right?
    f=-kx=-2.0E6*0.4=8.0E5N
    a=F/m=8.0E5N/1300Kg=615.4 kg.n
     
  11. Nov 23, 2009 #10
    If the spring is compressed by only 0.30-m, find the change in the mechanical energy due to friction.
    How do i find this part?
     
  12. Nov 23, 2009 #11
    KE= 1/2 2.0E6 N/m * (.3)^2
    Is this the right formula to use for the mechanical energy last part?
     
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