# Solving the Length of Ramp for a Box Weighing 200N

• perryrb
In summary, the box is lifted 2.0 meters using 350N of force and 76% of the work moves the box and 24% of the work overcomes the force of friction. The length of the ramp is 126.32 J due to the force of friction.
perryrb
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?

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.

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.

I still don't know the solution, but the answer in the book is 15.0 meters.

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).

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]

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 don't 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 don't 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 don't think ur book is right.

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!

## 1. How do you calculate the required length of ramp for a box weighing 200N?

To calculate the required length of ramp, you can use the formula L = h/(sinθ), where L is the length of the ramp, h is the height difference between the starting point and the end point, and θ is the angle of inclination of the ramp. In this case, h is the weight of the box (200N) and θ is the desired angle of inclination.

## 2. What is the ideal angle of inclination for a ramp when solving for the length?

The ideal angle of inclination for a ramp is typically between 15 and 20 degrees. This allows for a good balance between the required length of the ramp and the ease of pushing or pulling the object up the ramp.

## 3. Can the length of the ramp be determined without knowing the weight of the box?

No, the weight of the box is a crucial factor in determining the required length of the ramp. Without knowing the weight, it is not possible to accurately calculate the length of the ramp using the formula mentioned above.

## 4. How does the coefficient of friction affect the required length of the ramp?

The coefficient of friction between the ramp and the surface it is placed on can affect the required length of the ramp. A higher coefficient of friction would require a longer ramp to compensate for the increased resistance, while a lower coefficient of friction would require a shorter ramp.

## 5. Are there any safety precautions that should be considered when solving for the length of ramp?

Yes, it is important to ensure that the ramp is stable and secure, and that the surface it is placed on is level and free from any obstacles. It is also important to consider the weight and size of the object being moved and to use appropriate equipment and techniques to prevent injury.

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