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Friction block traveling problem

  1. Oct 21, 2007 #1

    bww

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    Please help,

    A 10kg block slides down a ramp from an elevation of 2 meters. At the bottom it encounters a rough surface, where a force of friction of 40 Newtons acts on it.
    a. How fast is the block traveling after it has crossed 2.5 meters of the rough patch?
    b. How much further will it travel?

    Thanks
     
  2. jcsd
  3. Oct 21, 2007 #2

    G01

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    Please show some work. It is the policy at this forum to not help with homework type questions unless some work is shown. Now no matter how lost you are, you still must know SOMETHING about this problem. Tell us how far you got, what you know, etc., so we can better help you.

    So, for starters how about telling us what you know, and your thoughts on how to get to a solution. You must have some thoughts!

    Also, please post homework type problems in the appropriate homework help section next time.
     
  4. Oct 21, 2007 #3

    bww

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    the answers are
    a. 4.3m/s
    b. 2.4m

    ...but, I have no idea how to arrive at these answers. Thanks.
     
  5. Oct 21, 2007 #4

    G01

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    Again, I repeat myself, the policy is that you must show some work or some form of independent thought on the problem. Me just giving you the answer won't help you understand or learn physics any easier.

    Again, again, I repeat myself. You must have some thoughts. Everybody has thoughts. What are your thoughts on this problem? What concepts do you think are involved?
     
  6. Oct 21, 2007 #5

    bww

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    ok, well the problem comes from a sheet of problems where i was solving potential and kinetic energy problems. So, this problem being different than those has thrown me off. I attempted to find the acceleration of the block before it hits the rough patch by using the PE formula of MxGxH (10kg x 9.8 x 2m) which gave me 196J. Then I tried using the KE formula of 1/2mv^2 and solved for v and got 6.3m/s. However, I'm not sure if I'm doing this right at all. It doesn't seem like the proper approach.
     
  7. Oct 21, 2007 #6

    G01

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    OK, this work is correct so far. Without you mentioning this, I had no idea where in the problem you were stuck. Now, I know it's when the block hits the rough patch and your work up to that point is correct. (Now, you see why it pays to show your work!)

    You have found the speed of the block before it hits the rough patch. Now, friction will slow it down, and you want to find the new speed 2.5 meters away. HINT: Think work. Are there any forces doing work on the block?
     
  8. Oct 21, 2007 #7

    bww

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    so when the block hits the rough patch it has a velocity of 6.3m/s. The frictional force of 40N acts against it, so it will slow down. Work is Fxd right? so a force of 40Nx2.5m=100. But what do I do with that number? Forces working on the block...gravity? so 9.8N pulling down....
     
  9. Oct 21, 2007 #8

    G01

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    OK, when we are talking about work, the only forces that do work are the forces in the direction of motion. So, since gravity is perpendicular to the direction of motion, it will do no work on the block, so we don't need to consider it. Now, that number you found, 100. It will have units of Joules right? So what is it's relationship to energy? What I'm asking you is, what is the relationship between work and energy?
     
  10. Oct 21, 2007 #9

    bww

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    ohh, the relationship between work and energy is W=KE+PE right?
     
  11. Oct 21, 2007 #10

    G01

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    KE+PE is the total mechanical energy of the system. Work on the other hand, is equal, here, to the change in mechanical energy of the system. So from the beginning of the rough patch to the end, only the kinetic energy changes. So, what does the work here equal then?
     
  12. Oct 21, 2007 #11

    bww

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    so if work equals the change in kinetic energy, the work done on the rough patch is 40Nx2.5=100J like before. That is the change in kinetic energy after the block travels 2.5m on the rough patch? so i solve for v in the KE formula again, I get 96=5v^2, divide 96 by 5, take the square root of that number and get v=4.38m/s. Cool. Now for part b....where do i start?
     
  13. Oct 21, 2007 #12

    G01

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    What will the energy be when the object is at rest? Can you figure out how much work needs to done in order to bring the object to this energy? If you can, you should be able to find the distance required.
     
  14. Oct 21, 2007 #13

    bww

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    well the potential energy will be 196J right? but the kinetic energy will be 0 when the object is at rest. so i use the difference between the PE and work (196-100=96J). That is my W, so 96J=40N(force of friction)d. 96/40 d=2.4m. Ok, so that works. Thanks a ton. Want to help me out on another problem while we're here....?

    A 1kg ball rolls along a circular track that is banked at 20 degrees. The normal force on the ball is 10.43N, tilted inward. The radius of the track is 10m.

    a. What is the net force on the ball (include direction)?
    b. How fast is the ball going for it to complete the turn?

    hmm this stuff is new i don't know too much, I know F=ma. The mass is 1 kg, and to find accelration I use v^2/r, right? but I don't know the velocity....
     
  15. Oct 22, 2007 #14

    G01

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    Draw a detailed diagram for this problem, with every force on it. You can find the direction of the net force by thinking about it. The object is moving in a circle, so what does this tell you about the direction of the net force on the object?

    When finding the strength of the net force, remember that in the direction of the net force:

    [tex]Net Force=\Sigma F = ma[/tex]
     
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