Coefficient Of Kinetic Friction Question

nintendude794

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
mass = m = 25 kg
angle = (theta) = 26 degrees
acceleration = a = 1.32 m/s²
gravity = F_g = 9.81 m/s²
coefficient of kinetic friction = M_k = F_k/F_n
F_k = resistance, friction? Force opposite Mass*Acceleration.
F_n = perpendicular to the object's motion, upward
3. The attempt at a solution
F_µ = F_n - 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. :)

Attached Thumbnails

 

ehild

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

F_k=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

nintendude794

Here, what exactly is "μ"? What does it mean?

ehild

It is the coefficient of friction. You used the notation M_k.

ehild

nintendude794

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. :)