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Conservation of Energy, PE & KE, block on incline with KE & friction?

  1. Oct 19, 2012 #1
    A 4 kg bundle starts up a 30° incline with 128 J of kinetic energy. How far will it slide up the plane if the coefficient of friction is 0.30?

    So the solution to the problem is 128 = 4(9.8)(x)(sin30) + 0.30 (4)(9.8)(x)(cos30)....x = 4.29m

    I dont understand the solution.

    so 128 = 4(9.8)(x)(sin30) + 0.30 (4)(9.8)(x)(cos30) is

    KE = mgh (sin theta) + Friction (mgh)(cos theta)

    I thought it would be KE = PE + Friction, why is mgh added twice???
     
  2. jcsd
  3. Oct 19, 2012 #2
    KE=PE-W where W is work done by friction.
    KE=mgh-Friction*distance
    Friction=normal force*coefficient of friction, now normal force is mgh*cos30
     
  4. Oct 19, 2012 #3
    The coefficient of friction is just a unitless factor that tells you how much the (maximum) force of friction compared to the normal force. What is the normal force?
     
  5. Oct 19, 2012 #4
    It really is KE = PE + Losses to friction

    have a closer look at the connection between the height above the starting point and how far along the slope the block travels
     
  6. Oct 19, 2012 #5
    Isn't friction just

    Coefficient x mg

    Without height?
     
  7. Oct 19, 2012 #6
    For friction, F=μFn. Fn isn't mg. Second, x isn't h. It's the distance the block slides along the surface. x sin(theta) is h. Third, x cos(theta) is also not h.
    What you are calculating is not the force, but the work, which is force times distance. The force has an mg term in it, but the work has an mgh term.
     
  8. Oct 19, 2012 #7
    Yes, it was a typo.
     
  9. Oct 19, 2012 #8
    By the way you have a little typo too. Fμ=μFn, now in y-direction ƩF=0⇔Gy=Fn →Fn=mg*cos30 So Fμ=μ*mg*cos30 Do you agree with me now?
     
  10. Oct 19, 2012 #9
    Ok that makes sense but why is the solution to the problem is:

    128 = 4(9.8)(x)(sin30) + 0.30 (4)(9.8)(x)(cos30)

    Which is
    KE = mgh (sin theta) + Friction (mgh)(cos theta)

    Shouldn't it just be
    KE = (mgh) sin theta + coefficient of friction(mg) (cos theta)????

    But that gives the wrong answer

    But it would make more sense because friction = mu x mg
     
  11. Oct 19, 2012 #10
    Were you talking about the part that looks like Fn*Fn? That's not a typo, it's two separate sentences.
     
  12. Oct 20, 2012 #11
    I am sorry, it really seemed to me Fn*Fn
     
  13. Oct 20, 2012 #12
    Look at the units.

    [KE]=J=kg*m2/s2

    [m]=kg
    [g]=m/s2
    [h]=m
    [sin(θ)]=unitless
    [mgh sin(θ)]=kg*m2/s2

    [μ]=unitless
    [m]=kg
    [g]=m/s/s
    [cos(θ)]=unitless
    [μmg cos(θ)]= kg*m/s2

    Without going into any physics, your answer is wrong because the units don't agree.


    Let's start from the top.
    1. What are the FORCES acting on the bundle? Write these in terms of their components parallel and perpendicular with the slope.
    2. What is the net force? It should be parallel to and down the slope.
    3. Use [itex]W=\int\vec{F}\cdot d\vec{x}[/itex] to calculate the WORKas a function of the distance the bundle travels. Remember, [itex]d\vec{x}[/itex] is in the direction parallel to and up the slope, so your answer should be negative.
    4. Solve KE+W=0 for x, the distance up the slope. This is NOT the distance in the horizontal direction, nor is it the height.
     
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