Calculating Friction and Velocity for a Slowing Object

Click For Summary

Homework Help Overview

The discussion revolves around calculating friction and velocity for an 85-N box of oranges being pushed across a horizontal floor while slowing down at a constant rate of 0.90 m/s². Participants are examining the forces acting on the box, including the horizontal and vertical components of the push force.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between acceleration and velocity, with some attempting to clarify the distinction between constant acceleration and constant velocity. There is also an exploration of the forces involved, particularly friction, and how they relate to the box's deceleration.

Discussion Status

The discussion is active, with participants correcting misunderstandings about acceleration and velocity. Some have provided clarifications regarding the units of acceleration and the implications of the box slowing down. There is an acknowledgment of the need to find the coefficient of friction, although no consensus on the approach has been reached.

Contextual Notes

Participants note potential confusion in the problem statement regarding the terms used for acceleration and velocity, leading to discussions about the logical consistency of the scenario presented.

Peach
Messages
80
Reaction score
0

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.
 
Physics news on Phys.org
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:
 

Similar threads

Replies
18
Views
3K
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
13K
  • · Replies 10 ·
Replies
10
Views
4K