Calculate the forces of a mover pushing a crate

  • Thread starter raman911
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  • #1
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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.



The Attempt at a Solution


Fg=mg
=85kg*9.8N/kg
Fg=833N
So Fn=Fg
but how can i find Fapp
 

Answers and Replies

  • #2
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You may want to review friction;

Fn=weight since it is vertical. But frictional force is a fraction of this. Hint:There is a constant that you are not using.
 
  • #3
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make a body diagram and it will be more clear. Net force of the forces in the direction of the displacement equal the product of mass and acceleration. Solve for Fapp..
 
  • #4
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Net force= forces applied on the axis of the displacement
 
  • #5
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Net force= forces applied on the axis of the displacement
i don't know net force
 
  • #6
arildno
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i don't know net force
Sure you do.
You know the crate moves with constant velocity.

What does that mean, in terms of net force acting upon the crate?
 
  • #7
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Sure you do.
You know the crate moves with constant velocity.

What does that mean, in terms of net force acting upon the crate?
i don't understand that
 
  • #8
arildno
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If an object moves with constant velocity, what is its acceleration?
And how is acceleration tied together with forces?
 
  • #9
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If an object moves with constant velocity, what is its acceleration?
And how is acceleration tied together with forces?
but i need time to find acceleration
 
  • #10
arildno
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  • #11
i need help with AP Physics free response questions
:frown:
 
  • #12
its the 2002 ap physics c queston
 
  • #13
arildno
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i need help with AP Physics free response questions
:frown:
Then make your own thread! :smile:
 
  • #14
could someone reply if they are willing to help
 
  • #15
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[tex]Given[/tex]

[tex]m=0.2kg[/tex]
[tex]\Delta {h}_{2}=0m[/tex]
[tex]{v}_{2}=3.96m/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*0m[/tex]
[tex]{E}_{g}=0J[/tex]


[tex]{E}_{k}=(1/2)m{v}_{2}^2[/tex]
[tex] =(1/2)0.2Kg(3.96m/s)^2[/tex]
[tex]{E}_{k}=1.56J[/tex]
 
Last edited:

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