Calculating Kinetic Friction

  • Thread starter Jim4592
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  • #1
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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
 

Answers and Replies

  • #2
18
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F=ma is your best friend :)
 
  • #3
49
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so using F=ma i would get....

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

Would that be a correct solution?
 
  • #4
18
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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=0


here 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....
 
  • #5
49
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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?
 
  • #6
18
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yes, you got it now!
 

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