Speed of a box at the bottom of a incline.

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To calculate the speed of a box traveling down an incline with a 22-degree angle and a kinetic friction coefficient of 0.22, the user initially misapplied the energy conservation equation, assuming a frictionless scenario. The correct approach involves accounting for the work done by friction, which reduces the box's kinetic energy. The user successfully derived the final velocity using a kinematics equation after clarifying the role of friction and normal force. It was emphasized that the kinematics equations are valid only under constant acceleration conditions. Understanding the impact of friction is crucial for accurate speed calculations at the bottom of the incline.
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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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