Coefficient Of Kinetic Friction Question

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Homework Help Overview

The discussion revolves around a physics problem involving a 25 kg box sliding down a 26-degree ramp with a specified acceleration. Participants are tasked with finding the coefficient of kinetic friction between the box and the ramp, utilizing given values such as mass, angle, and gravitational acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the forces acting on the box, including the normal force and kinetic friction. There are attempts to express these forces mathematically, with some questioning the notation used for the coefficient of friction.

Discussion Status

Some participants have provided insights into the forces involved and the equations that relate them, while others express confusion about specific terms and concepts. The original poster indicates a lack of understanding and has sought assistance, suggesting that the discussion is ongoing without a clear resolution yet.

Contextual Notes

The original poster mentions a tight deadline for the assignment, which may influence the urgency and nature of the discussion. There is also an indication of varying levels of understanding among participants, with some needing clarification on fundamental concepts.

nintendude794
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Homework Statement


A 25 kg box slides down a 26 degree ramp with an acceleration of 1.32 m/s2.
The acceleration of gravity is 9.81 m/s2
Find the coefficient of kinetic friction between the box and the ramp.

Homework Equations


mass = m = 25 kg
angle = (theta) = 26 degrees
acceleration = a = 1.32 m/s2
gravity = Fg = 9.81 m/s2
coefficient of kinetic friction = Mk = Fk/Fn
Fk = resistance, friction? Force opposite Mass*Acceleration.
Fn = perpendicular to the object's motion, upward

The Attempt at a Solution


Fµ = Fn - COS26(I don't know...)

Attached is a pic of my teacher's process for solving the same problem, only with different values.

I'm clueless. Teacher said this is the toughest of the 14 problems that are due in 6 hours, so I thought I might seek assistance in learning how to do it. Please and thanks. :)
 

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The force of kinetic friction is proportional to the normal force Fn, and its acts along the slope in opposite direction as the body moves. Fn, the normal force is the component of mg which is perpendicular to the slope,

Fn=mgcos(theta).

The force of kinetic friction is

Fk=μ mg cos(theta).

The component of mg which is parallel with the slope is

Fp=mg sin(theta),

and it points down the slope. The force of kinetic friction points in the opposite direction. The resultant of these forces will accelerate the body down the slope

ma=mg sin(theta)-μ mg cos(theta).

Given m, a, theta, g, find μ.

ehild
 
ehild said:
The force of kinetic friction is

Fk=μ mg cos(theta).

ehild

Here, what exactly is "μ"? What does it mean?
 
It is the coefficient of friction. You used the notation Mk.

ehild
 
ehild said:
It is the coefficient of friction. You used the notation Mk.

ehild

Thanks. This assignment was due 30 minutes ago, I accepted a zero if even only temporarily because I do not understand. I have printed it out to take it to tutorials. Thanks so much for trying to help me though. :)
 

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