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Homework Help: How to get coefficient of f^k without mass and without constant force

  1. Nov 25, 2011 #1
    1. The problem statement, all variables and given/known data
    When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 1.99 m/s. The mass stops a distance S2 = 2.5 m along the level part of the slide. The distance S1 = 1.18 m and the angle θ = 27.70°. Calculate the coefficient of kinetic friction for the mass on the surface.

    2. Relevant equations

    *maybe relevant*


    f^k = (Normal Force) * uk

    mechanical energy = (potential energy: h*m*g)+(kinetic energy: 1/2m*v^2)

    the general idea that kinetic friction is going to take away from the total energy.

    3. The attempt at a solution

    My first attempt actually involved kinematic equations, however i soon learned that this was a silly route (i will need to know at least "m" if i were to solve it this way...) So then i tried with mechanical energy...

    Potential energy = h(which is 1.18sin(27.7 deg) * m * g(which is 9.8m/s^2) = 5.375m

    Kinetic energy = 1/2m*v(which initially is 1.99)^2 = 1.98m

    so E_mech,i = (7.355m)J and E_mech,f = 0

    This also made me realize that i chose the wrong method...

    Then i decided to analyze f^k a little further. I know it has a similar relationship to f = ma. This way i can effectively get rid of this whole "m" thing. However, I don't have any evidence of what the object is doing after it exits the slope. All i know is that it gains x amount of velocity going down the 27.7 deg slope in 1.18m, then it takes 2.5m of f^k to slow it down to a stop.

    I then decided to try it out as a sort of work problem. I also decided to draw a straight line between x_i and x_f.....

    So yeah I am obviously not getting the point here. I think there is some basic info in this question that i am supposed to be paying attention to, but i'm not. This is a problem in a section where we talk about energy, power, and work. The whole thing where the object "m" suddenly goes from a theoretical acceleration due to the Normal Force minus f^k to a pure decelleration of f^k is really giving me hell. I know just about nothing about this object's movement, all i really know is that at some point it moved from the top of that slope to the x_f point in an unknown period of time.
    Last edited: Nov 25, 2011
  2. jcsd
  3. Nov 25, 2011 #2


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    Staff: Mentor

    The figure you refer to doesn't show up. Can you upload it?
  4. Nov 25, 2011 #3
    ok, does it show now?
  5. Nov 25, 2011 #4


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    Staff: Mentor

    Yes it does. Thanks.

    It looks like conservation of energy is the way to approach this. You're given two distances and you should be able to find expressions for the frictional force for each section (leave the mass as a variable, M. It'll cancel out eventually). When the mass comes to rest, all of its kinetic energy will have been given up to work done by friction.
  6. Nov 25, 2011 #5
    thanks for the direction, I am going to try this one again.
  7. Nov 25, 2011 #6
    okay so here goes

    So i figured out that kinetic energy is the spotlight here, i can disregard potential energy because it's a useless value in this equation.

    here are my variables:

    ki = starting kinetic energy = 1.98(mass)
    kf = final kinetic energy = 0
    k(s1) = kinetic energy at the end of distance "s1" = ki + ((fg-f^k)*s1)
    fg = positive force on object during distance "s1" (in direction of its motion) = 9.8(mass)*sin(27.7deg)
    f^k(s1) = kinetic friction in the direction that opposes motion during distance s1 = 9.8(mass)*cos(27.7deg)*uk
    f^k(s2) = kinetic friction opposing motion during distance s2
    s1 = distance1=1.18m
    s2 = distance2=2.5m

    So first i took the kf = ki + w equation and made kf be the point at the end of s1 because i want to split this up into s1 and s2 so i can single out the friction force.

    k(s1) = ki + ((fg-f^k)*s1)

    then i took the formula again and...

    0 = k(s1) + w
    that turns into

    -w = k(s1)
    now to make it bigger and uglier...
    -(f^k(s2)*s2) = ki + ((fg-f^k)*s1)
    and uglier
    -(f^k(s2)*s2) = 1.98m + ((9.8(mass)*sin(27.7deg)-9.8(mass)*cos(27.7deg)*uk)*s1)
    and uglier
    -(9.8m*uk*s2) = 1.98 + ((9.8(mass)*sin(27.7deg)-9.8(mass)*cos(27.7deg)*uk)*s1)

    EVENTUALLY I get that uk = 6.055... which is the wrong answer.

    I am pretty sure there is some way easier solution to this problem. I can't even find problems like it online! I almost wonder if i keep misreading this problem, or just don't have enough basic experience with energy conservation to be able to tackle it.

    Any help would be rad.
  8. Nov 25, 2011 #7


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    Staff: Mentor

    I think you're dissecting the problem too finely as you try to work through the minutia step by step. Let's take a step back. In the broadest picture you've got sources of energy and sinks for energy. At the start there are two sources for energy: The initial kinetic energy of the object and the initial potential energy of the object. At the end all the available energy from the sources has been given up to work done by friction.

    So start by adding up your assets. What's the initial total available energy?
  9. Nov 25, 2011 #8
    The total avaliable energy, or as I have been informed as "mechanical energy" is 7.355(mass)J

    ui = (1.18*sin(27.7deg)*9.8m)J or (5.375*mass)J
    ki = (1.98*mass)J
    So now that I look at it this way, I suppose that the important function s1 serves is that that is the period of which all of the potential energy turns into kinetic energy

    s2, then, would be the period of which all of the kinetic energy then dissipates.

    I've been looking at my notes and I think I will now try a changing mechanical energy model: Ef = Ei + W where:
    Ef = final mechanical energy = 0
    Ei = starting mechanical energy = (7.355*mass)J
    W = work done by a non conservative force such as friction

    0 = (7.355*mass)J + W
    that would mean
    -W = (7.355*mass)
    w = force*distance, and i am going to assume the distance, since it is the work of friction, is s1+s2=3.68m
    Also F is the kinetic friction force which is 9.8*mass*uk
    -(9.8*mass*uk*3.68) = (7.355*mass)J
    now i will divide mass
    uk = -0.203
    Now i am going to assume that i messed up a sign somewhere and that uk actually = 0.203...


    ok so the answer IS uk=0.203!

    Thanks for your help!
  10. Nov 25, 2011 #9


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    Staff: Mentor

    Not to harsh your buzz or anything :smile: but you should keep the division between the two sections of the trajectory since they will each experience a different amount of energy loss per unit length. Note that S1 is along a slope so it will have a lower normal force (to cause friction) than the horizontal section S2.

    On S2 the normal force is just Mg, so the friction there is μkMg. On the slope it'll be different. You can still equate the starting mechanical energy to the energy lost to friction, but you'll need to take these two sections into account.

    The energy lost on the horizontal section is F*d = μkMgS2. What is it on the slope?
  11. Nov 25, 2011 #10
    if energy lost on the horizontal section is F*d = ukMgS2
    I suppose on the incline the normal force would be 9.8M*cos(27.7deg) which means f^k would be 9.8M*cos(27.7deg)*uk

    so the energy lost on the incline is 9.8M*cos(27.7deg)*uk*S1?

    and that just leaves the energy gained on S1, which would be (5.37*mass)J which is what was the potential energy prior. And, of course, the initial kinetic energy of (1.98*mass)J and I think that leaves me with enough to solve the problem.
  12. Nov 25, 2011 #11


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    Staff: Mentor

    Yup. So the total energy lost on both sections is:
    [tex] \mu_k M g S_2 + \mu_k M g cos(\theta)S_1 = \mu_k M g (S_2 + cos(\theta) S_1)[/tex]
    Equate this to your initial mechanical energy and you're home free.
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