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Conservation of energy of sliding block

  1. Jul 28, 2005 #1
    Hey folks,

    I was trying to work out a problem on conservation of energy and am totally stuck. Was hoping someone could help.... the problem was this...

    A 38 kg block slides with an initial speed of 7 m/s up a ramp inclined at an angle of 35o with the horizontal. The coefficient of kinetic friction between the block and the ramp is 0.5. Use energy conservation to find the distance the block slides before coming to rest.

    I was working with the assumption that at the end of the climb,
    PE = KE + Wf
    where, PE = potential energy
    KE = kinetic energy
    Wf = work done by friction = coefficient * Normal force

    Reason I get stuck is that there is no d (displacement on slope) or h (height of block at the end of motion). I tried solving one for the other by using h/d = sin theta.... but I don't think it's right. :cry:

    ******EDITED PART STARTS*******
    Here's what i had done:
    PE = KE - Wf (minus cuz going up the slope)
    mgh = mv2 - (F*d) (F = friction_coeff * m * g * cos theta)

    dividing both sides by m*g*d I got....
    h/d = (v2/dg) - (F/mg)

    but, h/d = sin theta
    so,
    v2/dg = (F/mg) + sin theta

    then I solved for d... and got an answer of 5.1 m. However this was wrong!
    ******EDITED PART ENDS******

    And I've spent an hour on this and am totally frustrated and thus useless ... any and all assistance/clue/guidance will be totally appreciated.

    Thanks.

    - O
     
    Last edited: Jul 28, 2005
  2. jcsd
  3. Jul 28, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I would write it like this:
    [itex]{KE}_i + {PE}_i = {KE}_f + {PE}_f + W_{friction}[/itex]

    The work done by friction will equal [itex]\mu N d[/itex], where d is what you are trying to find, the distance the block slides.

    That relationship between h and d is exactly correct. Now make use of it in the energy equation and solve for d. (Hint: Express the PE in terms of d.)
     
  4. Jul 28, 2005 #3
    made an addendum

    Hey Doc Al

    I was editing my previous post when you posted your reply. As you'll notice from my edit of the original post... I think I did what you were asking me to do. I got that answer... but found out that it was wrong.

    As for the equation you wrote :
    [itex]{KE}_i + {PE}_i = {KE}_f + {PE}_f + W_{friction}[/itex]

    Both [itex]{PE}_i[/itex] and [itex]{KE}_f[/itex] are zero... right? That is why I used the equation I had. Am i wrong?

    Thanks again.
     
    Last edited: Jul 28, 2005
  5. Jul 28, 2005 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Careful: [itex]{KE} = 1/2 m v^2[/itex]. (You are missing a 1/2.)
     
  6. Jul 28, 2005 #5
    OMG... I can't believe I did that!!! :biggrin: .. .and to think I wasted an hour because of a 0.5!! I'll try working it out when I get home and see if it's correct.

    Thanks a heap!!.
     
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