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Block sliding down a ramp problem

  1. Nov 25, 2007 #1

    ~christina~

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

    The path that a block slides along is frictionless until the block reaches the section L shown in the figure below. The length L = 0.75 m begins at height h = 2.0 m on a ramp of angle θ = 30º. In that section, the coefficient of kinetic friction is 0.40. The block passes through point A with a speed of 8.0 m / s.

    If the block can reach point B (where the friction ends), what is its speed there, and if it cannot, what is its greatest height above A?

    [​IMG]

    2. Relevant equations

    [tex]K_A = U_g + K_f -f_kd [/tex]

    [tex]f_k=\mu_kmg [/tex]


    3. The attempt at a solution

    L (friction spot)= 0.75m
    [tex]h_f= 2.00m [/tex]

    [tex]\theta= 30^o [/tex]

    [tex]\mu = 0.40 [/tex]

    [tex] V_A= 8.0m/s [/tex]


    I was thinking conservation of momentum...

    do I take A. to be the initial V? I was thinking it would be but not sure. (continuing as if I'm correct)

    [tex]K_A + U_g= U_g + K_f -f_kd [/tex]

    [tex]0.5 mv_A^2 + 0 = mgd sin 30^o + 0 - \mu mgL [/tex]

    Is this correct?
    or does the distance for the friction have to include the angle?
    and thus be

    [tex]0.5 mv_A^2 + 0 = mgy sin 30^o + 0 - \mu mgLsin 30^o [/tex]
     
    Last edited: Nov 26, 2007
  2. jcsd
  3. Nov 25, 2007 #2

    rl.bhat

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    Find the velocity of the block at h = 2m. Then use the conservation of energy to find the final velocity at B.
    While going up the block encounters retardation due gravity and frictional force. While calculating the final velocity take these in account. If you get negative answer then the block won't reach the point B.
     
    Last edited: Nov 25, 2007
  4. Nov 25, 2007 #3

    ~christina~

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    was my equation correct though ? (to find the height)

    I was really curious to know if I needed to make the frictional force include the angle I was given.

    [tex]0.5 mv_A^2 + 0 = mgd sin 30^o + 0 - \mu mgLsin 30^o [/tex]
     
    Last edited: Nov 25, 2007
  5. Nov 25, 2007 #4

    rl.bhat

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    Frictional force = mu*mg*cos(theta). So the total work done along the ramp = [mu*mg*cos(theta) + mg*sin(theta)]*L
     
  6. Nov 25, 2007 #5

    ~christina~

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    why is it all multiplied by L ?

    L= only the part of ramp with friction.

    d= total distance traveled by block
     
  7. Nov 26, 2007 #6

    rl.bhat

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    Because the frictional force acts only along L, not through out the displacement. Since we don't know the total distance d try to work out with h and L. In your equation work done against the friction should be mu*mg*cos(theta)*L.
     
  8. Nov 26, 2007 #7

    ~christina~

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    [tex]0.5 mv_A^2 + 0 = mgy sin 30^o + 0 - \mu mg L cos 30^o [/tex]

    is it fine now?

    well assuming it is since nobody answered me.

    WAIT A SEC I don't have the mass of the block...Does it cancel out??

    Mass matters so how would I find it if the mass isn't given??
     
    Last edited: Nov 26, 2007
  9. Nov 26, 2007 #8

    rl.bhat

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    WAIT A SEC I don't have the mass of the block...Does it cancel out??
    Yes.
    Now what is y?
     
  10. Nov 26, 2007 #9

    ~christina~

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    y= 2m

    I figured that for the first part where I go and calculate the final v at the height 2m

    How would I calculate the v at the final point B ? I'm not quite sure how the eqzn would look.


    [tex]0.5(m)(8.00m/s)^2 + 0 = (m)(9.81m/s^2)(2m) + 0.5(m)(v_f)^2 [/tex]

    I think this is fine to calculate the v at the 2m point on the ramp but I'm not sure.

    I got v= 4.98m/s

    but how would I find the v at B (to see whether it's negative or not?)


    Is it

    [tex]0.5 mv_A^2 + 0 = mgy sin 30^o + 0 - \mu mg L cos 30^o [/tex]


    I have to get to sleep...I'll check in the morning
     
    Last edited: Nov 26, 2007
  11. Nov 26, 2007 #10

    rl.bhat

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    Height of B with respect to A ...H = h + l = 2 + 0.75sin(30).
    Now you can wright 0.5mVA^2 - 0.5mVB^2 = mgH + mu*mg*cos(30)*L.
    Solve for VB.
     
    Last edited: Nov 26, 2007
  12. Nov 26, 2007 #11

    ~christina~

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    so when I wrote this => [tex]0.5(m)(8.00m/s)^2 + 0 = (m)(9.81m/s^2)(2m) + 0.5(m)(v_f)^2 [/tex]

    I guess it was okay right? (I was thinking I needed a sin or cos somewhere)

    for the

    I don't have m so how would that work? I don't think it will cancel out this time.

    Help
     
  13. Nov 26, 2007 #12

    rl.bhat

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    I have edited my mail which you have quoted.
    The correct expression is: 0.5mVA^2 - 0.5mVB^2 = mgH + mu*mg*cos(30)*L.
    When you cancell m, you get...0.5*8^2 - 0.5*VB^2 = 9.8*(2 + 0.375) + 0.4*9.8*0.866*0.75. Solve it and find the value of VB.
     
  14. Nov 26, 2007 #13

    ~christina~

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    for the initial velocity at [tex]V_{2.00m}[/tex] I got 4.98m/s with the equation

    [tex]0.5(m)(8.00m/s)^2 + 0 = (m)(9.81m/s^2)(2m) + 0.5(m)(v_f)^2 [/tex]

    why did I get that if I don't use that as a initial velocity for V at h = 2.00 but instead use the VA which is bellow ramp to find V at the point B?
     
    Last edited: Nov 26, 2007
  15. Nov 26, 2007 #14

    rl.bhat

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    Initially I suggested that keeping in mind to use the formula:
    0.5*V2m^^2 - 0.5*VB^2 = 9.8* 0.375 + 0.4*9.8*0.866*0.75. After seeing your equation I thought it is better to do at one stretch. In your equation the error is you have assumed VB = 0 and you have taken work done against friction negative.
     
  16. Nov 26, 2007 #15

    ~christina~

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    Okay Thank You :smile:
     
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