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Homework Help: Grade 11 Physics: Solving for Distance with Friction, Energy, and Variables

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data

    The top of a hill is height h, m above the ground. If there is a friction patch on the flat ground only and no friction on the hill, what will the speed of a box of mass m at the bottom of the hill? How long must the friction patch be to make the box's speed 1/3 of what it was at the bottom of the hill if the coefficient of friction between the box and the patch is [tex]\mu[/tex]?

    2. Relevant equations

    Ek=1/2mv^2, Eg=mgh

    3. The attempt at a solution
    I've already solved the first part of the question in regards to the speed. For the first part, I got [tex]\sqrt{}2gh[/tex] as the answer for v2. However, I'm stuck at the second part. So far, the attempt is:

    v1=0 h2=0
    Et[tex]\neq[/tex] Ek
    1/2mv1^2=1/2m(v2/3)^2
    0=1/2m(v2/3)^2
    And from then on, I don't really know what to do.
     
  2. jcsd
  3. Apr 27, 2010 #2

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    Hello Felicity26,

    From the time the box hits the bottom of the hill, until the time the velocity is slowed to 1/3 its initial value (where the initial value is at the time the box hits the bottom of the hill), the box loses kinetic energy. How much kinetic energy is lost? Where does this energy go? :wink:
     
  4. Apr 27, 2010 #3

    kuruman

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    OK, so you know that the speed of the box is sqrt(2gh) when it hits the friction patch. You are also told that its speed is sqrt(2gh)/3 when it exits the friction patch of length, say, x. Can you think of a kinematic equation that relates all these quantities?
     
  5. Apr 27, 2010 #4
    I don't think I can use sqrt(2gh) for the second part because the mass is at the bottom of the hill and the height is zero, so that would make the entire thing zero. Should I assume that the deacceleration is constant? Because if that's the case, then I can use v2^2=v1^2+2ad.
     
  6. Apr 27, 2010 #5

    kuruman

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    The symbol "h" represents the height of the hill, not the position of the box. If the hill is 20 m high, then h = 20 m, regardless of whether the box is at the top of the hill or at the bottom.
    Yes you should and yes you should use the equation you quoted.
     
  7. Apr 27, 2010 #6

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    Keep in mind what your results from the first part mean. You have calculated that the box's speed is [tex] \sqrt{2gh} [/tex] at the very bottom of the hill (and this speed is non-zero). Even though the speed might be written in terms of variables, it is really just a number. So given this number for the speed, what kinetic energy is associated with it (in therms of the same variables)?

    Perhaps you can answer this one. :tongue2: What is the equation governing the force of dry friction on a flat surface?

    (Hint: Once you have the above, instead of invoking kinematics, you might find it easier to use a different relationship between energy and distance. But then again, the kinematics should work too) :wink:
     
  8. Apr 27, 2010 #7
    So h actually doesn't matter?
     
  9. Apr 27, 2010 #8

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    h matters very much! :smile: But as Kuruman says, h represents the height of the hill, not the height of the object at any given time.

    Think of it this way, if the height of the hill is larger, then the box will have a faster speed when it hits the bottom. How fast will it be going? The answer is a function of h. So its speed, and thus its kinetic energy (at the bottom of the hill) can be expressed in terms of h.

    And by the way, your final answer, the distance d, that the object travels across the rough surface, will also be expressed in terms of h (as well as some other variables such as possibly g, [tex] \mu [/tex], m, etc).
     
    Last edited: Apr 27, 2010
  10. Apr 27, 2010 #9
    Alright, I get that. But can you elaborate a little more about the dry friction equation, because I really have no clue what that is. So far, I've only been taught at school about the Ek=1/2mv^2 equation anda couple of really basic ones. Thanks :)
     
  11. Apr 27, 2010 #10

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    I encourage you to look in your textbook and/or course material for the specific equations I'm about to describe. I'm sure they are in there. I'll describe them in words to point you in the right direction.

    The magnitude of the force of kinetic friction on an object is equal to the coefficient of kinetic friction (between the object and the surface) times the magnitude of the normal force applied to the object by the surface. I.e., you should be able to represent the force of kinetic friction in terms of the kinetic friction coefficient and the normal force.

    Work is a measure of energy (or more appropriately, a measure of overall change in energy). For a constant force, the work done on a system is the vector dot product of the force and the displacement over which the force is applied. Because it is a vector product, there may be cosθ term that fits in, if the force and displacement are not parallel to each other. For the simple case where the force and displacement are parallel, the work is simply equal to the magnitude of the constant force times the magnitude of the displacement.

    Yes, that might sound a little wordy. But you should be able to find more concise versions, in equation form, in your textbook or coursework.
     
  12. Apr 27, 2010 #11
    Oh, I see what you're talking about. It's that Ff=Mu(Fn) and W=F(d). So then I find the friction force and then use that to calculate the distance. I see.
     
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