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Incline problem with friction; alternative solution

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data
    There is an example problem in a text book I'm looking at where they solve a simple incline-mass problem (friction included) using work-energy.

    We are given the mass 40kg, the length of the slide is 8 m, the incline is 30 degrees, the coefficient of kinetic friction is .35. We are looking for the child's speed at the bottom of the slide.

    2. Relevant equations
    [tex] W_{ext}=\Delta E_{mech}+f_k \Delta x [/tex]

    Newton's Seconds Law, the basic kinematic equations:

    3. The attempt at a solution

    The solution in the example is straight forward. The child-slide-earth system has no exterior forces acting so the equation I give above is set to zero.

    [tex]0=mg\Delta h +\frac{1}{2}mv_f^2+f_k \Delta x=-mg\Delta x sin(\theta)+\frac{1}{2}mv_f^2 +mg\mu_k cos(\theta)\Delta x[/tex]

    Plugging in numbers and solving for the final velocity they obtain:
    [tex]v_f \approx 5.6 m/s [/tex]

    I am attempting the same problem usuing Newton's second law and kinematic equations, but I am not obtaining a similar answer and I can't discern why not:

    In the x-direction, using the usual coordinate system (x - axis parallels to slope of the incline, +x is down the incline), the acceleration should be:

    [tex]a_x = gsin(\theta) -g \mu_k cos(\theta) [/tex]

    Plugigng in numbers real fast gives:
    [tex]a_x \approx 1.93 m/s^2[/tex]

    Now I attempt to use kinematics to find the speed at the bottom of the slide:

    [tex]x=\frac{1}{2}a_x t^2 [/tex]

    [tex]v_f=a_x t [/tex]

    [tex]80m = \frac{1}{2}(1.93 m/s^2)t^2\Rightarrow t\approx 9.11s [/tex]

    Using this in the velocity equation along with the acceleration of 1.93 m/s^2 gives about 17.6 m/s for the speed. This is far different than the 5.6 m/s obtained using the energy equations.The numbers I obtained don't pass sanity check either.
     
  2. jcsd
  3. Oct 15, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    In your last equation you've used 80 m as the length of the slide. Wasn't the value only 8 m in the problem statement?
     
  4. Oct 15, 2016 #3
    You gotta be kidding me. I can't believe I didn't notice that and wasted this much time. I guess I'm kind of relieved that it was my vigilance and not my understanding of physics that failed though. I thought I was going crazy for moment.
     
    Last edited: Oct 16, 2016
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