Calculate the forces of a mover pushing a crate

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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
 
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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.
 
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..
 
Net force= forces applied on the axis of the displacement
 
chouZ said:
Net force= forces applied on the axis of the displacement

i don't know net force
 
raman911 said:
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?
 
arildno said:
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
 
If an object moves with constant velocity, what is its acceleration?
And how is acceleration tied together with forces?
 
arildno said:
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
 
raman911 said:
but i need time to find acceleration

Eeeh, why?
 
i need help with AP Physics free response questions
:frown:
 
harmonicmotion said:
i need help with AP Physics free response questions
:frown:

Then make your own thread! :smile:
 
could someone reply if they are willing to help
 
[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]
 
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