Speed of a box at the bottom of a incline.

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    Box Incline Speed
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Homework Help Overview

The discussion revolves around calculating the speed of a box at the bottom of an incline, given an angle of 22 degrees, a height of 9.10 m, and a coefficient of kinetic friction of 0.22. The original poster, Paul, is attempting to determine the speed after the box descends the incline from rest.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of gravitational potential energy and the impact of friction on the calculations. There are questions about the use of specific equations and the role of mass in determining forces. Some participants suggest reviewing energy conservation principles and the effects of nonconservative forces.

Discussion Status

The discussion is active, with participants exploring various equations and concepts related to energy and friction. Some guidance has been provided regarding the need to account for friction in the calculations, and there is acknowledgment of the limitations of certain equations in this context.

Contextual Notes

Participants note the challenge of calculating forces without knowing the mass of the box, which is a critical factor in applying the equations discussed. There is also a recognition that the original approach may not yield correct results due to the presence of friction.

Paulbird20
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Ok so I am having trouble calculating the speed of a box as it travels down a incline.

I was Given 22 degree angle and the coefficient of friction (kinetic) is .22. Height of 9.10m (from rest at the top).
I calculated acceleration while it travels down the slop and got 1.6721.
I next need to calculate the speed when it reaches the bottom of the incline.

I thought i would simply need to use 2*G*H ^(1/2) . It gives me 13.355 and it is wrong.
Any help would be greatly appreciated or any equations i can use.

Thanks from a new user.
Paul.
 
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Why not use the approximation of gravitational potential energy for objects close to Earth?

U[g]=mgh

But that doesn't just equal the kinetic energy at the bottom. You have to take into account the work done by nonconservative forces, namely, in this case, friction.

I don't know if that was any help or not.
 
I don't know how to calculate the mass to be able to use that equation. Only equations i have in my notes are
2*G*H^(1/2) = V1
and
calculations of sum of all forces. I have a free body diagram drawn also i just don't know where to go from here. I was almost sure 13.355 was gunan be the speed but it is wrong.
 
That equation is true if the block is frictionless but it is not frictionless in this case. Go back and re-read your notes or the book to better understand the U[g]=mgh gravitational potential energy equation.

The equation that you have comes from conservation of energy:

(mv^2)/2=mgh; rearrange that
v=(2gh)^1/2

But, as I said before, you cannot apply that equation here because the surface is not frictionless. Some energy is lost to friction.

Here's a hint: W[friction] =F[friction]*distance

Subtracting this value from your initial potential energy will give you the final kinetic energy at the bottom of the block.
 
Ok so from what i have gathered reviewing my notes.

Fnormal= m*g*cos (angle)
and
FF(friction force) = coef K * Fn

I was trying to use these two equations to determine FF but i can't because of the mass in the first one. Is there another equation i can use to get FF?
 
AH HA! i found my equation your hint helped thank you.

I took 2 * A* x1-x0 ^(1/2) and it gave me my final velocity Thank you.
 
Ah, I see. You used a kinematics equation. Just remember, it may have worked in this case but those five major equations only work when the acceleration is a constant value.
 

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