Calculating Friction and Velocity for a Slowing Object

AI Thread Summary
An 85-N box of oranges is pushed across a horizontal floor, slowing at a constant acceleration of 0.90 m/s². The forces acting on the box include a horizontal push of 20 N and a vertical component of 25 N downward. The discussion clarifies that the box's deceleration indicates a non-zero acceleration, contradicting the notion of constant velocity. The participants emphasize the importance of correctly interpreting the problem's units and the relationship between friction and acceleration. Ultimately, the goal is to determine the coefficient of friction based on these forces and the box's deceleration.
Peach
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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


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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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.
 
Okay firstly, I need to correct your question. Notice the alteration is red
Peach said:

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.
Peach said:
If it's slowing down, that's only because of the friction force right?
Correct
Peach said:
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 :rolleyes:
 
But the unit is m/s, isn't acceleration m/s^2?

Edit: Okay nvm, I got it. Thanks for both the help!
 
Peach said:
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:
There's no typo and the units given in the problem statement are correct:
Peach said:
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.)
 
Doc Al said:
There's no typo and the units given in the problem statement are correct:
Note that it says 0.90 m/s each second, which is equivalent to 0.90 m/s/s or m/s^2.
:redface:
#Crawls and hides in shame#
:blushing:
 
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