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Spring launching 3 kg block

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data
    A horizontal spring with a spring constant of 200.0 N/m is compressed 25.0 cm and used to launch a 3.00 kg block across a frictionless surface. After the block travels some distance, the block goes up a 32 degree incline that has a coefficient of friction of 0.25 between the block and the surface of the incline. How far along the incline does the block go before stopping?



    2. Relevant equations
    motion equations
    f=-ks
    f=ma


    3. The attempt at a solution
    i tried taking the force of the spring at .25m compression which is 50N times sin32 to account for incline i got 26.5N, minus the force of the friction which i got as 6.23 so a total force of 20.27 divided by the mass gives acceleration of 6.76. Then i was not able to use a motion equation to find a distance up the incline
     
  2. jcsd
  3. Nov 17, 2011 #2
    When the mass is on the incline there is no component of the force of the spring anymore. The effect of the spring appears only in the velocity of the mass up the plane.

    Which forces do you think act on the mass while it is moving up the incline?
     
  4. Nov 17, 2011 #3
    force of friction and gravity?
     
  5. Nov 17, 2011 #4
    There is also the normal reaction of the incline on the mass.

    Now which method are you going to use, energy or equations of motion?
     
  6. Nov 17, 2011 #5
    okay
     
  7. Nov 17, 2011 #6
    it would be mgsin32 i believe
     
  8. Nov 17, 2011 #7
    the normal reaction is mgcos32.
     
  9. Nov 17, 2011 #8
    so normal incline is mgcos32 = 24.93 N and ff = .25mg =7.35N so total resistance along the incline is 32.28N?
     
  10. Nov 17, 2011 #9
    normal reaction = 0.25x24.93=6.23N
     
  11. Nov 17, 2011 #10
    frictional force = coefficient of friction x normal reaction
     
  12. Nov 17, 2011 #11
    thats what i got originally, how do i use that to find a distance up the ramp?
     
  13. Nov 17, 2011 #12
    are you going to use energy or equations of motion?
     
  14. Nov 17, 2011 #13
    im guessing energy because i didnt see a way of doing it with motion equations?
     
  15. Nov 17, 2011 #14
    so initial energy = final energy
    what is formula for the initial energy?
    what is formula for final energy?
     
  16. Nov 17, 2011 #15
    initial energy = Us=1/2kX^2 = 1/2(200)(.25)^2 = 6.25 KE=1/2mv^2 6.25=1/2(3)V^2
    Vi= 2.04?
     
  17. Nov 17, 2011 #16
    correct.
    In what form is the final energy?
     
  18. Nov 17, 2011 #17
    Kinetic Energy with a velocity 0? or maybe potential energy cause its not moving?
     
  19. Nov 17, 2011 #18
    if v = 0 there is not ke
    So final energy is mgh.

    Hence can you write an equation showing what happened to the intial energy?
     
  20. Nov 17, 2011 #19
    so 6.25=mgh so h= .213m ?
     
  21. Nov 17, 2011 #20
    6.25J has to provide energy for the final PE (mgh) and ALSO for the work done against friction.
     
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