What am I missing?

  • Thread starter perryrb
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  • #1
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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?
 

Answers and Replies

  • #2
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perryrb said:
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?

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.
 
  • #3
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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.
 
  • #4
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I still don't know the solution, but the answer in the book is 15.0 meters.
 
  • #5
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perryrb said:
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.

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).
 
  • #6
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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]
 
  • #7
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perryrb said:
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]

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.
 
  • #8
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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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