What is the coefficient of friction for a pen rolling down a book?

AI Thread Summary
To find the coefficient of kinetic friction for a pen rolling down a book, it is essential to understand that the net force acting on the pen is the difference between the gravitational force component parallel to the slope and the frictional force opposing the motion. The pen accelerates at 3.28 m/s², and the normal force is 0.161 N, which corresponds to the perpendicular component of the pen's weight. The correct approach involves using Newton's second law, where the net force equals mass times acceleration, and considering both the parallel and perpendicular components of the weight. It's important to note that for calculating the coefficient of kinetic friction, the pen must be sliding rather than rolling. The discussion emphasizes the need to clearly define the forces acting on the pen to derive the correct equations.
cs44167
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Homework Statement
If a pen accelerates down a book at 3.28 m/s/s, what is the coefficient of kinetic friction between the pen and book?
Relevant Equations
from earlier part: Normal force = 0.161 N
mass of pen = 18.7 g
perpendicular component of weight = 0.161 N
parallel component of weight = 0.0869 N
angle of book = 28.3 degrees
I tried using coefficient of friction = friction / Normal force, but needed a value for friction. I then tried to find the friction using a = f/m, but was unsure of which value to plug in for force. Simply finding the force given a and m will not yield the correct answer; the net force must be a difference of two forces.

Any suggestions?
 
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cs44167 said:
Homework Statement: If a pen accelerates down a book at 3.28 m/s/s, what is the coefficient of kinetic friction between the pen and book?
Homework Equations: from earlier part: Normal force = 0.161 N
mass of pen = 18.7 g
perpendicular component of weight = 0.161 N
parallel component of weight = 0.0869 N
angle of book = 28.3 degrees

the net force must be a difference of two forces.
Actually the net force is the sum of all the forces some of which may be negative. Anyway, can you think of two separate forces actng on the pen? If so, write an expression for the net force and set it equal to mass times acceleration. Be sure to use symbols and do not substitute numbers until the very end. BTW, if you are looking for the coefficient of kinetic friction, the pen must be sliding, not rolling as the title of this post suggests.
 
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kuruman said:
Actually the net force is the sum of all the forces some of which may be negative. Anyway, can you think of two separate forces actng on the pen? If so, write an expression for the net force and set it equal to mass times acceleration. Be sure to use symbols and do not substitute numbers until the very end. BTW, if you are looking for the coefficient of kinetic friction, the pen must be sliding, not rolling as the title of this post suggests.

I know the weight acts on the pen directly down, and the friction acts on the pen up the slope of the incline. I also know the components of the weight, where the parallel component is parallel to the slope, and the perpendicular component in perpendicular to the incline and has the same magnitude as the Normal force. I’m just not sure how to input which of these into an equation.
 
If we take the x-axis to be parallel to the slope (in the same direction as that of the velocity or the acceleration of the pen) what equation can you make use Newton's 2nd law for this axis?. You going to use only the x-components of the forces on this x-axis.
 
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