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Pushing object up incline - no friction

  1. Jun 7, 2007 #1
    1. The problem statement, all variables and given/known data

    What is the minimum work needed to push a 1200Kg car, 118m up 18 degree incline with no friction?

    2. Relevant equations

    I used mgsin18 - mgcos18 (don't think this is right)

    I know with friction, you do mgsin18 - (coef. friction)(mg)(cos18), and then to get work you multiply that answer by the distance. But without friction it doesn't work the same way, but I'm not sure what to do.

    3. The attempt at a solution
    using mgsin18 - mgcos18, I get a negative answer which is wrong. The correct answer is 4 x 10^5, but I get -7290. WAY off!
  2. jcsd
  3. Jun 7, 2007 #2
    draw a free body diagram showing all the forces on the block. (and never memorize any of this kind of situation dependent equations)
    by the way mgsinx - mgcosx is wrong
    assuming that there is no friction (because the constant is not given)
    using newton's rules find the minimum force needed
    use the definition of work and given info. to calculate work.
  4. Jun 7, 2007 #3
    As a side thought to the problem, will the fact that the block is moving up a frictionless "incline" make the work done less than when lifted vertically up to the same height?
  5. Jun 8, 2007 #4
    i think that the min force that should be applied is mgsin 18 since you are doing work against the component of weight parallel to the inclined slope only. that's it.... and if i'm not wrong it gives a work done of 4.38 X 10^5 J

    eeerrmmm.... for the side thought... i believe that moving the block up the incline plane will require less work than lifting it up to the same height.not considering air resistance)

    coz... up the plane, only a component of the weight is acting, while when lifting it up, the whole of the weight acts.

    hope iwas right
  6. Jun 8, 2007 #5

    Doc Al

    User Avatar

    Staff: Mentor

    But work is force times displacement. How does the displacement change between those two scenarios? Figure out the work done in each case.
  7. Jun 8, 2007 #6

    eeerrmmm.... for the side thought... true, you are working against force weight*sin(theta) but the displacement has also increased to height/sin(theta).
    product of these two still remains same.
    in both cases work done is same
  8. Jun 8, 2007 #7
    up the incline, displacement is the hypotenuse
    and for lifting it right up, displacement is the vertical side of the right angle triangle

    hehe...hurray....i have got 50 posts in PF
  9. Jun 8, 2007 #8
    and for the friction thing,
    put μ = 0....what else??
  10. Jun 8, 2007 #9
    that means no friction
    therefore there is no energy lost to friction
    ascending the object to the same hight or moving it up the incline require the same amount of work

    (conservation of energy)
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