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What am I missing?

  1. Oct 20, 2006 #1
    I'm having difficulty with this problem.

    A box weighing 200N is lifted 2.0 meters up a ramp using 350N. 76% of the work moves the box and 24% of the work overcomes the force of friction between the box and the ramp. What is the length of the ramp?

    Work in the y-axis is conserved = 200N x 2.0 meters = 400 J
    Work due to friction is not conserved but can be calculated
    Total Work (x and y) = 400 J / 0.76 = 526.32 J
    Work due to friction is 126.32 J
    Length of ramp = 126.32 J / Force of friction
    Length of ramp = 126.32 J / Uf x 200N x cos(angle between gravity and normal of the ramp)

    Is this correct? How do I calculate the length of the ramp?
  2. jcsd
  3. Oct 20, 2006 #2


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    I just get the feeling that the 2 m is the vertical height above the ground to which the box is lifted. So, you can surely find the potential energy of the box at a height of 2m, right? Where does that P.E. come from? Now 24% work is wasted. leave that. So, the rest 76% of your work is spent in lifting the box. So, u can easily form the equation using Work done= Force x displacement. Obviously the displacement is the length.
  4. Oct 20, 2006 #3
    2.0 meters is the height, so P.E. = mgh = 200N x 2.0m = 400J
    This is where I'm confused. The answer is 15.0 meters.
  5. Oct 20, 2006 #4
    I still don't know the solution, but the answer in the book is 15.0 meters.
  6. Oct 20, 2006 #5


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    Ya, obviously the answer will be 15m. And where did the g go? its 4000 J, not 400. Now 76% of the total work done by you will be 4000. Now, solve that. In PF, u'll have to do some work by yourself, no one will post full solutions (PF rules).
  7. Oct 20, 2006 #6
    I don't mind the work, I've been working at this, I appreciate the response!

    P.E. = mgh = mg x h = 200N{mg} (weight of box) x 2.0m{h} = 400J [N x m]
  8. Oct 20, 2006 #7


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    Then i dont think it'll ever come to 15. maybe the answer in the book is wrong or there is a misprint in the question. I didnt notice that it was the weight of the box that is 200. I thought it was the mass. if u take it to have a mass of 200 kg and not weight, then the answer is 15m. Other than that, i dont know. U'll have to wait for someone who has a better knowledge of physics, to reply to this thread. maybe he can say something. But i dont think ur book is right.
  9. Oct 20, 2006 #8
    Thanks, yeah I think it's an error in the text. It does work out if the box has a mass of 200kg.

    [200kg x 9.8m/s^2 x 2.0m / 0.76] / 350N = 15m

    This isn't the first error I've encountered in a REA study book. Again, Thanks for your help!
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