Finding distance traveled up a ramp

In summary, the conversation discusses solving for the distance traveled by an object using the equations ΣFΔt=mΔv and ΣF=ma. The speakers share their attempts at solving the problem and discuss using both x and y components of weight in the calculations. They eventually reach the correct solution of 1310.90 meters traveled.
  • #1
x2017
84
1

Homework Statement


Untitled.png


Homework Equations


ΣFΔt=mΔv
ΣF=ma
v=d/t

The Attempt at a Solution


I already found the following for previous questions:
mg=-661.9
mgx=-3594.37
mgy=-649.91
Fx=2192.69
Fy=-1250.69
FF=-365.44
Vf=257.18
ΣFx=-1767.12

ATTEMPT #1
v=d/t
d=vt
d=(257.18)(10)
d=2571.80 (incorrect)

ATTEMPT #2
ΣFΔt=mΔv
ΣFΔt=m(d/t)
[ΣFΔt/m](t)=d
[(-1767.12)(10)/67.4]10=d
-2621.84=d

ATTEMPT #3
ΔV=Δd/Δt
ΔvΔt=Δd
(257.18-5)(10)=Δd
2521.80=Δd

I am not sure where I am going wrong!
 
Last edited:
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  • #2
Your x and y components of the weight are greater than the weight.
 
  • #3
TSny said:
Your x and y components of the weight are greater than the weight.
IMG_0105.JPG


I keep getting the same things (Sorry about the bad quality photo, but I figured this was best so the triangles I made can be seen).
 
  • #4
In your pink triangle drawn on the inclined plane, note that mg is the hypotenuse. But in the work shown in the blue rectangle you have mgx as the hypotenuse.
 
  • #5
TSny said:
In your pink triangle drawn on the inclined plane, note that mg is the hypotenuse. But in the work shown in the blue rectangle you have mgx as the hypotenuse.

If I use the pink triangle again for mgx and use cos I get mgx=121.63.
But I'm confused because I still got 3594.37 for mgy & that's much larger than mg (661.19) in the first place, so is that one incorrect?

EDIT: mgy has to be correct because I had to calculate the force of friction in a previous question with it and obtained the correct answer!
 
  • #6
x2017 said:
But I'm confused because I still got 3594.37 for mgy & that's much larger than mg (661.19) in the first place, so is that one incorrect?

EDIT: mgy has to be correct because I had to calculate the force of friction in a previous question with it and obtained the correct answer!
In the pink box you got the correct value for mgy.

Your value for mgx now looks correct also.
 
  • #7
TSny said:
In the pink box you got the correct value for mgy.

Your value for mgx now looks correct also.

Okay, thanks!

So am I supposed to use both when solving for distance or just the stuff in the x direction? I was only using the stuff in the x direction before.
 
  • #8
You need to use the y direction information to help get the friction force. But once you have the friction, you just need to consider ΣFx = max.
 
  • #9
TSny said:
You need to use the y direction information to help get the friction force. But once you have the friction, you just need to consider ΣFx = max.

Okay so:

ΣFx=max
Fx-FF-mgx=max
2192.69-365.44-121.63=67.4[(vf-vi)/Δt]
(1705.62/67.4)10=vf-vi
253.06=Δv

And this is where I get lost... I am stuck solving for a distance, I don't know how to proceed.
 
  • #10
Never mind, I figured it out!

d=[(vi+vf)/2]t
d=[(5+257.18)/2]10
d=1310.90m
 

Related to Finding distance traveled up a ramp

What is the formula for finding distance traveled up a ramp?

The formula for finding distance traveled up a ramp is d = h / sin(θ), where d is the horizontal distance traveled, h is the vertical height of the ramp, and θ is the angle of the ramp.

What units of measurement should be used for the height and distance in the formula?

The height and distance should be measured in the same unit, such as meters or feet. It is important to use consistent units in order to get an accurate result.

How does the angle of the ramp affect the distance traveled?

The angle of the ramp plays a significant role in the distance traveled. As the angle increases, the distance traveled also increases. This is due to the fact that a steeper ramp requires more force to overcome the gravitational force, resulting in a longer distance traveled.

Are there any factors that can affect the accuracy of the calculation?

Yes, there are a few factors that can affect the accuracy of the calculation. These include the smoothness of the ramp, the surface of the ramp, and any external forces acting on the object, such as friction or air resistance.

Can this formula be applied to all types of ramps?

Yes, this formula can be applied to all types of ramps, as long as the ramp has a constant angle throughout its length. If the angle of the ramp changes, the formula would need to be modified accordingly.

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