Calculate the work done by the mover

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Homework Help Overview

The problem involves calculating the work done by a mover pushing a crate along a floor, considering the effects of friction and the forces acting on the crate. The subject area includes dynamics and work-energy principles.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of the crate moving at constant speed, questioning the net forces acting on it. There are attempts to clarify the meaning of "constant" and its relation to acceleration. Some participants express confusion about the calculations related to the applied force and the frictional force.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's calculations and questioning the reasoning behind specific values. There is no clear consensus yet, as participants are exploring different interpretations and clarifying concepts.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information shared and the methods discussed. There is a focus on understanding the relationships between forces rather than arriving at a final answer.

raman911
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1. Homework Statement
A mover pushes a 85kg crate along the floor at a constant speed through a displacement of 3,1m[E]. The coefficient of kinetic friction between the floor and crate is 0.22.
A) Calculate Fn and Fapp.
B] Calculate the work done by the mover.3. The Attempt at a Solution
Fg=mg
=85kg*9.8N/kg
Fg=833N
So Fn=Fg
but how can i find Fapp

B] w=183N*3.1m
w=567.3
 
Last edited:
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The box is moving at a constant speed, what does that say about the sum of the forces on the box?
 
please helpp....
 
What does the word constant mean??
 
arildno said:
What does the word constant mean??

not changing
 
Yes, so what is the acceleration of the box? How can you use that to find the sum of all the forces?
 
daniel_i_l said:
Yes, so what is the acceleration of the box? How can you use that to find the sum of all the forces?
So a=0
Fnet=0
Fnet=Fapp+Ff
Fapp=0-180
Fapp=180
is that right?
 
Last edited:
[tex]Given[/tex]

[tex]m=0.2kg[/tex]
[tex]\Delta {h}_{1}=0.835m[/tex]
[tex]{v}_{1}=0m/s[/tex]

[tex]Required[/tex]

[tex]{E}_{g} and {E}_{k}[/tex]

[tex]Solution[/tex]

[tex]{E}_{g}={m}g\Delta h[/tex]
[tex]={0.2kg}*9.8N/kg*0.835m[/tex]
[tex]{E}_{g}=1.63J[/tex][tex]{E}_{k}=(1/2)m{v}_{1}^2[/tex]
[tex]=(1/2)0.2Kg(0m/s)^2[/tex]
[tex]{E}_{k}=0J[/tex]

[tex]{E}_{T}=1.56J[/tex]
 
Last edited:
no response
 
  • #10
Where did you get the 180 from, and why did you cut out the sign in front of it?
 
  • #11
Your working seems ok, but I don't know where you got 180 from. Check your math...

Edit: Dammit, that's three times that's happened today
 
  • #12
Hootenanny said:
Your working seems ok, but I don't know where you got 180 from. Check your math...

Edit: Dammit, that's three times that's happened today

soory i mean 183
 
  • #13
I couldn't care less if the number was 54 or square root of pi fifths.

I want you to tell us:

1. HOW did you arrive at that number??

2. WHY did you take away the sign?
 

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