How to Find Coefficient of Kinetic Friction?

AI Thread Summary
To find the coefficient of kinetic friction in this scenario, consider the energy dynamics involved as the wagon moves up and down the incline. The potential energy at the top of the hill is converted into work done against friction as the wagon slides down and travels on the flat surface. The normal force on the incline is affected by the weight of the wagon and the incline angle, necessitating a careful breakdown of forces. The problem can be approached in two parts: analyzing the motion on the incline and then on the horizontal surface. Understanding these energy transformations and forces will lead to calculating the coefficient of kinetic friction accurately.
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Homework Statement



A man is pulling a wagon up a 50° up a hill. His nephew of in unknown mass sits in the wagon. After pulling the wagon 30 meters up the incline, the man slips and falls releasing the wagon. The sled, starting from rest, slides down the incline. After traveling 30 meters the sled travels 50 meters on a wide open horizontal surface before finally coming to a stop. Assuming the coefficient of friction is the same throughout the entire problem, find the coefficient of kinetic friction. If the coefficient of static friction between the wagon and the ground was twice the coefficient of kinetic friction, would the wagon move at all?

Hint: Treat this as two different parts with two different coordinate systems.

Hint: The speed the wagon reaches at the bottom of the hill is the dame speed the wagon starts with when traveling on the ground.

Homework Equations



Idk?

The Attempt at a Solution



I don't know where to start on this problem :confused:. Can you guys help at all?
 
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Wow, an interesting one! Welcome to PF, Physics Girl.
I'm still wondering if it is a wagon or a sled.
It is clearly an energy problem because of the PE and friction.
The big picture is
PE at the top of the hill = work done against friction
and there is the friction going down the hill and the friction along the flat to be done separately. Each is basically a W = F*d. The F is the force of friction. On the hill, the normal force is not the weight, but the component of weight pushing normal (perpendicular) to the hill.
See what you can do with it; start with the big picture and fill in the details! You will certainly need a diagram of the hill - put the mass right at the top; makes it easier to draw the mg down and show its normal and parallel components forming a triangle with the full mg.
 
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