How to Calculate Coefficient of Kinetic Friction for Slowing Box?

[Jim4592]

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

 

[jmarcian]

F=ma is your best friend :)

 

[Jim4592]

so using F=ma i would get...

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

Would that be a correct solution?

 

[jmarcian]

on which axis is the box accelerating, (or deccelerating)?
does the box have any y-direction movement?

think about those, and remember
\sumF=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...
\sumF_x=ma_x, and
\sumF_y=ma_y
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,
\sumF_x=ma_x, and
\sumF_y=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...

 

[Jim4592]

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?

 

[jmarcian]

yes, you got it now!