How to Calculate Coefficient of Kinetic Friction for Slowing Box?

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Homework Help Overview

The problem involves calculating the coefficient of kinetic friction for an 85-N box being pushed across a horizontal floor while decelerating at a rate of 0.90 m/s². The forces acting on the box include a horizontal push force of 20 N and a vertical component of 25 N downward.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law (F=ma) and consider the forces acting on the box in both horizontal and vertical directions. Questions arise regarding the acceleration direction and the relevance of the vertical force component.

Discussion Status

The discussion is ongoing, with participants exploring the relationship between the forces and acceleration. Some guidance has been offered regarding the breakdown of forces into components and the importance of drawing a free body diagram (FBD). There is a recognition of the need to clarify the acceleration's direction and its implications for the forces involved.

Contextual Notes

Participants are navigating the constraints of the problem setup, including the horizontal movement of the box and the implications of the vertical force component. There is an emphasis on understanding the forces acting on the box in the context of its motion.

Jim4592
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Homework Statement


An 85-N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward.


Homework Equations


Calculate the coefficient of kinetic friction between the box and the floor.


The Attempt at a Solution


I have no idea how to solve this problem, here's what i wrote down on the test and only missed 7 points on a 17 point problem:

A = -0.90m/s^2

Fx = µxN
Fx = -0.90 * 20
Fx = -18 N
 
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F=ma is your best friend :)
 
so using F=ma i would get...

F = 85N * -0.90 m/s
F = -76.5N

Would that be a correct solution?
 
on which axis is the box accelerating, (or deccelerating)?
does the box have any y-direction movement?

think about those, and remember
[tex]\sum[/tex]F=ma, this means all the forces on the same axis are equal to the mass of the body times its acceleration..

breaking F=ma into components we get...
[tex]\sum[/tex]Fx=max, and
[tex]\sum[/tex]Fy=may
on which axis is the box accelerating? is there any acceleration in the y-direction? if you read the question it says the box is being pushed across the floor horizontally which means no y-acceleration, do you see that? so now we have,
[tex]\sum[/tex]Fx=max, and
[tex]\sum[/tex]Fy=0here is your part now draw out your FBD, your second best friend, and ALL the forces acting on it... then decide which force goes in which equation...
 
so we can just ignore the vertical component of the force?

and get:

20 N - Friction Force = 85N * A(x)

and would Ax be -0.90 m/s?
 
yes, you got it now!
 

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