# Friction and velocity help

## 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.

Basically, f=ma

## The Attempt at a Solution

Okay, this is what I attempted. I drew a FBD, so the eqn I came up with is
fk - Fx = 0
fk = Fx
fn - w - Fy = 0
fn = w + Fy
And since fk = uk(fn)
Then shouldn't uk = Fx/(w+Fy)?

I know that's wrong because I didn't include the velocity. But I don't know how to integrate the velocity in there. If it's slowing down, that's only because of the friction force right? And at a constant rate just means the acceleration is zero....What other logic am I missing? Many thanks in advance.

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Doc Al
Mentor
What you are given is not a velocity, but an acceleration. It's undergoing a constant acceleration, not constant velocity. The sum of forces in the horizontal (x) direction is not zero, since the box is accelerating.

Hootenanny
Staff Emeritus
Gold Member
Okay firstly, I need to correct your question. Notice the alteration is red

## 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/s2 each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward.
If the box is slowing then it is undergoing an acceleration, which is the rate of change of velocity wrt time and has units of meters per second per second. I'm going to ignore your working, because I think you have a fundamental misunderstanding here.
If it's slowing down, that's only because of the friction force right?
Correct
And at a constant rate just means the acceleration is zero
This is simply wrong. If an object is slowing down how can the acceleration possibly be zero? Remember, as I said above that acceleration is the rate of change of velocity. It therefore follows that if the box is slowing down it has a non constant velocity (i.e. its velocity is changing) and hence the acceleration must be non-zero. Do you follow?

EDIT: Dammit Doc But the unit is m/s, isn't acceleration m/s^2?

Edit: Okay nvm, I got it. Thanks for both the help!

Hootenanny
Staff Emeritus
Gold Member
But the unit is m/s, isn't acceleration m/s^2?
You're correct, so I assume either you've incorrectly copied the problem or there has been a typo. However, you should have considered this yourself, the question as written is illogical; how can a box be slowing down and yet have a constant velocity?

Last edited:
Doc Al
Mentor
There's no typo and the units given in the problem statement are correct:
As it moves, it is slowing at a constant rate of 0.90 m/s each second.
Note that it says 0.90 m/s each second, which is equivalent to 0.90 m/s/s or m/s^2.

(Of course you did leave out part of the problem statement: The question itself! Which I assume is to find the coefficient of friction.)

Hootenanny
Staff Emeritus  