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Inclined plane work

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    A block of mass m1 is placed on an inclined plane with slope angle a and is connected to a second hanging block of mass m2 by a cord passing over a small frictionless pulley the coefficient of kinetic friction is uk the system is released from rest with block m2 dropping a distance h.

    given m1, a, h, m2, uk
    calc the work due to each of the forces on block 1 in moving h
    calc the work due to each of the forces on block 2 in moving h
    calc the total work on the sys and final speed of the blocks

    2. Relevant equations

    w = F*d

    3. The attempt at a solution

    http://i.imgur.com/IxtPw.png

    for block1

    Wg = -h*mg*sin a

    Wt1 = hT

    Wn=0

    Wff = -h*uk*mg*cos a

    Wtot = hT - *mg*sin a - h*uk*mg*cos a


    for block2

    Wg = mgh

    Wt = -hT

    Wtot = mgh - hT


    hT - *mg*sin a - h*uk*mg*cos a + mgh - hT = 1/2*m*Vf^2 + 1/2*m2*Vf^2 - 0 (start from rest)

    sqrt( 2/(m1+m2) * (mg*sin a - h*uk*mg*cos a + mgh)) = Vf



    ----

    blocks a, b, c are connected as shown. block a,b,c have same mass m and coefficient of kinetic friction between each block and the surface is uk. block c descends with constant velocity. use 30 for angle of incline. given m h uk determine

    work due to forces on A moving h
    work due to forces on B moving h
    work due to forces on C moving h
    solve for uk in terms of the other givens for the system to move at constant speed



    http://i.imgur.com/DtwxJ.png


    block A

    Wff= -uk*mg*h

    Wt= T1*h

    block B

    Wt2 = T2*h
    Wg = -mg*sin a*h
    Wff= -uk*mg*cos a * h
    Wt1 = -T1*h

    wtot = T2*h - mg*sin a * h - uk*mg*cos a * h - T1* h

    block C

    Wg = mg*h
    Wt2 = -T2*h

    Wtot = mg*h - T2*h


    T1*h - uk*mg*h + T2*h - mg*sin a * h - uk* mg* cos a * h - T1*h + mg*h - T2*h = 0(constant speed implies vf v0 are the same meaning no change in kinetic energy right??)

    -uk*mg*h - uk*mg*cos a * h = mg*sin a * h - mg*h

    -uk - uk*cos a = sin a - 1

    -uk(1 + cos a) = sin a - 1
    uk = - (sin a - 1/ (1+cos a))

    uk = 1/2 / (2+sqrt(3))/2

    uk = 1/(2+sqrt(3))






    want to be sure im not doing anything horribly wrong thanks for any help.
     
  2. jcsd
  3. Oct 12, 2012 #2

    SammyS

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    Wow! A little bit of punctuation, etc. sure helps ! ...along with a few other changes.
     
  4. Oct 14, 2012 #3
    didnt seem to be an issue on other posts, sorry
     
  5. Oct 14, 2012 #4

    SammyS

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    Do you mean, in your other posts?

    In many of those you did, at least, use periods.

    You probably should take a bit of time to read rules for posting in this Forum. Here are two short excerpts.
    General Posting Guidelines
    ...Pay reasonable attention to written English communication standards. This includes the use of proper grammatical structure, punctuation, capitalization, and spelling. ...

    Do not hijack an existing thread with off-topic comments or questions--start a new thread.
     
  6. Oct 14, 2012 #5
    are you having trouble interpreting what I've typed? clearly not or you wouldn't be capable of correcting it by replacing one word and telling me to use more periods. I think that fulfills reasonable english communication.

    now read the last sentence of your post. do you not see the irony?
     
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