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Newton's Laws, find the time it takes to reach the top of the ramp. Mechanics.

  1. May 13, 2010 #1
    Hey all, here's a question from one of my homework sets, any help is greatly appreciated.


    1. The problem statement, all variables and given/known data

    I've included a diagram of the situation. Basically, I have to find out how long it takes the block to reach the end of the ramp, the spring is massless, and friction is a factor in this problem.

    edit: Mass of block is .1KG, not 100KG. The picture does not show this very clearly.
    phys_example.jpg



    2. Relevant equations

    Newton's Second Law: F = MA

    Potential energy of a spring: Us= .5KX^2

    Friction: F = kN



    3. The attempt at a solution

    I know that to get started in this problem, I will first need to find the magnitude of each force acting on the block, and in turn find the net force. My reference frame is such that +X lies parallel to the ramp in the direction the block is moving, and +Z is the direction perpendicular to the ramp.

    Here's my attempt to find the forces acting in the +X direction (at the instant before the Block loses contact with the spring):

    Gravitational,
    I know that a portion of the gravitational force will be resisting the block's movement, but I'm running into my first problem here. I found the total gravitational force to be MG = .1kg * 9.8m/s2 = .98N. However, I'm having problems setting up a triangle to find the ratio of how much the .98N is in the +X direction, since the force is not running parallel with the hypotenuse. How do I find it?

    Spring Energy,

    I know that the spring constant is 400N/m, and it is compressed .1m.

    U2 = .5*400N/m*(.1m)^2 = 2.0J

    So the speed just after launch by conservation of energy is:

    2.0J = .5*.1kg*v2
    V = 6.32m/s

    However I'm not sure if this is entirely useful, as I'd like to know the force on the block, how do I find this if I don't know the acceleration?

    Friction, I know friction on an incline depends on the normal force, so:

    Fkf = .250*.1kg*9.8m/s2*cos(10°) = .241N in the -X direction.





    I think I have the force of friction correct, so I'm 1 for 3 so far. If anyone could help me out with finding the gravitational force and the force for the spring I'd greatly appreciate it.
     
    Last edited: May 13, 2010
  2. jcsd
  3. May 13, 2010 #2

    diazona

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    Actually, you already did. It's in this calculation:
    Also, remember Hooke's law? (The relationship between spring force and spring compression)
     
  4. May 13, 2010 #3
    Ah, so the .241N is the total force of gravity + kinetic friction opposing the motion of the block?



    I had forgotten about Hooke's law, let's see if I have it right:

    FSpring= kx so,

    F = 400N/m * .1m = 40N


    Thanks for the quick reply.
     
  5. May 13, 2010 #4

    diazona

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    No, not at all - what you calculated there is just the force of kinetic friction, since you used the formula for kinetic friction. The component of gravity doesn't magically get included.

    Let me ask you this: what was your thought process in coming up with the way you calculated the kinetic friction? e.g. what equation(s) did you use?
     
  6. May 14, 2010 #5
    I know that kinetic friction depends on the Normal force of the object, and that in this case the normal force would only be a ratio of the total amount of gravitational force [MGcos(10°)*coefficient of friction]...you got me, I think just found my logical error. So the total amount of force resisting motion is :

    Kinetic friction: .241N

    +

    Gravitational force: MGcos(10°) = .965N

    =

    1.206 in the -X direction.



    So, if I was correct when finding the spring energy, the net force in the X direction is

    40N(spring) - 1.206N(friction + gravitational) = 38.794N
     
  7. May 14, 2010 #6

    diazona

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    Actually no, that's not it...
    Where does that come from? How did you know to write MG cos(10°)?
     
  8. May 14, 2010 #7
    Well, I didn't know for sure, but I assumed that because I wanted the total force of gravity to be on the hypotenuse of the right triangle used to find the x and z components of gravity, I adjusted the triangle's position accordingly. And now that I'm taking a second look at my work this morning after getting some sleep, I see that it should be MGSIN(10°).
     
  9. May 14, 2010 #8

    diazona

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    And that's exactly the answer to your first question, i.e. how you find how much of the gravitational force acts in the +x direction.
    Which component are you talking about here, the x or the z component?
     
  10. May 14, 2010 #9

    Z component, correct?
     
  11. May 14, 2010 #10

    diazona

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    No, check that triangle again.
     
  12. May 14, 2010 #11

    Okay, I see that it is in the X-direction.
     
  13. May 14, 2010 #12

    diazona

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    OK... by the way, I think I mixed up the components myself at some point in there, so sorry if I misled you :blushing: But I'm fairly positive it's right now.

    So, now you know how to find the component of gravity that acts against the motion. You've also found the frictional force. Now, what about the spring force?

    The aim is to find the total force acting on the mass. Then you can apply Newton's second law to find its acceleration.
     
  14. May 14, 2010 #13
    Haha yeah you did stump me for a while there, thanks for all the help you've given so far, I appreciate it.

    So, recapping I have these forces acting on the block in the X direction:

    Force of gravity: .1kg * 9.8m/s2*cos(10°) = .965N -X

    Force of Kinetic Friction: .250*M*G*cos(10°) = .241N -X

    Force of Spring: 400N/m * .1m = 40N +X


    Net force in X direction = 38.794N

    By Newton's Second Law acceleration is:

    38.794N/.1kg = 387.94m/s2


    Does everything look good up to this point?
     
  15. May 14, 2010 #14

    diazona

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    Hold on, remember what you said a few posts back:
    If you're confused about whether it should be a sine or cosine, what I like to do is imagine the angle being 0° (or 90°) and see whether your formula makes sense.

    Here's something else to think about: in the force of gravity formula (the previous line in your post), you'll have a factor of either sin(10°) or cos(10°). In this formula for the force of kinetic friction, you'll have a factor of either sin(10°) or cos(10°). Should you have the same trig function in both formulas, or should it be sin(10°) in one and cos(10°) in the other?

    This applies when the spring compression is .1m. What about when the spring compression is less than .1m? (when it's partially uncompressed?)

    Also, keep in mind that there are two regions of motion to consider, the part while the spring is pushing the block and the part after the block has left the spring. You'll need to figure out the time for each one separately and add them up.
     
  16. May 14, 2010 #15
    I think I'm finally seeing the light, when finding the gravitational force I need to use sin(10°), and when finding the the kinetic friction force, I need to use cos(10°), as kinetic friction uses a ratio of the normal force, which is in the Z direction.


    I'm not quite sure how to account for that, do I need the conservation of energy equation (Us = 1/2*kx^2), to find the speed at a certain point?

    Yes, I do realize this, but I'm not 100% sure how to account for it, as we don't know the acceleration during decompression, only the speed after decompression, and the distance traveled.

    Alright, I'm hoping I finally have the net force correct:

    Gravitational: .1Kg*9.8m/s2*sin(10°) = .170N -X

    Kinetic Friction: .250*.1Kg*9.8m/s2*cos(10°) = .241N -X

    Spring (at the instant before it loses contact with block) = 400N/m*.1kg = 40N +X

    Net Force = 40N - .241 - .170 = 39.59N +X

    Acceleration = 39.59N/.1kg = 395.9 m/s2
     
  17. May 15, 2010 #16

    diazona

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    Yeah, that's it.
    Ordinarily, yes, that's exactly what you could do. But in this case it's kind of tricky, because both friction and gravity are still in effect, even while the spring is decompressing. The presence of those other forces affects both the speed of the block when it loses contact with the spring, and the position at which it loses contact with the spring. (It's not quite .1m from the starting position, that distance is slightly modified) So for this one I'd suggest that you stick with forces and Newton's second law.
    Well, that you can figure out.

    Actually... I should check: have you studied simple harmonic oscillators?
    Seems okay. And you can write these forces as -.170N and -.241N, rather than explicitly writing "-X" after each one. It will make the calculations easier to represent the forces using positive and negative numbers.
    That one still needs some work. You need to use Hooke's law, which says that the force exerted by the spring is proportional to the spring's compression (or extension) from its equilibrium position. Here are some questions to get you started:
    - If you ignore friction and gravity (for the moment), at the very beginning of the problem, what is the spring's displacement from equilibrium?
    - Still ignoring friction and gravity, when the block loses contact with the spring, what is the spring's displacement from equilibrium?
    - Now start to consider friction and gravity again. Can you tell how that modifies the answers to the last two questions? (Hint: one of them changes, the other does not)
     
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