Calc Push Force Req'd to Move 13g Steel Block on Steel Table

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Homework Help Overview

The discussion revolves around calculating the work required to push a 13g steel block across a steel table with a given coefficient of friction at a steady speed. Participants are exploring the relationship between force, friction, and work in the context of physics principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of force using F=MA and the implications of steady speed on net force. Questions arise regarding the calculation of friction force and its relationship to the applied force.

Discussion Status

The discussion has progressed with participants confirming that the net force is zero due to constant velocity, leading to the conclusion that the pushing force equals the friction force. Some calculations have been attempted, and guidance has been provided regarding the relationship between forces.

Contextual Notes

Participants are working under the constraints of a homework problem, focusing on the definitions and relationships between force, friction, and work without reaching a definitive conclusion on the final answer.

bjbaldwi
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How much work must you do to push a 13g block of steel across a steel table(coefficient of friction=.6) at a steady speed of 1.2 for 8.8 ?

I know that Work=Force*Distance

i know what the D is but i don't know how to calculate the force.
i tried F=MA and got Fnet= Force pushed -Friction force
i really have no i dea tho. I need help please.
 
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bjbaldwi said:
i tried F=MA and got Fnet= Force pushed -Friction force
Hint: What does 'steady speed' tell you about the net force?

How do you calculate the friction force?
 


Steady speed says the friction is constant and friction force is kinetic friction*normal force or Fk*mg mg= normal force i think
 


OK, good. But what does Fnet equal? (What's the acceleration of the block?)
 


sorry its 13 kg 1.2 m/s and 8.8 s
 


the acceleration is zero because it is a constant velocity
 


bjbaldwi said:
the acceleration is zero because it is a constant velocity
Good. So what's Fnet? And what does that tell you about Fpush?
 


Fnet is equal to zero since the acc. is zero. F push is equal to friction force?
 


bjbaldwi said:
Fnet is equal to zero since the acc. is zero. F push is equal to friction force?
Exactly! You should be able to calculate the work now.
 
  • #10


Fpush is equal to 0 and f net is equal to friction force

so Fnet=Fpush+Ffriction
Fpush=0
Fnet=Ffriction
Ffriction=Fnet =(.6)mg---->.6*13*9.8=76.44
so
Work=Fnet*D
D = 1.2m/s*8.8s---->10.56 m
so
76.44*10.56=807.2 J
 
  • #11


Thanks for all your help : )
 
  • #12


bjbaldwi said:
Fpush is equal to 0 and f net is equal to friction force
No, you had it right before. Fnet = 0 so Fpush = Ffriction.

so Fnet=Fpush+Ffriction
Fpush=0
Fnet=Ffriction
No, but that doesn't affect your calculation, which is correct.

It's the force Fpush that is doing the work.
 

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