Finding Friction Force & Work with No Applied Force

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Homework Help Overview

The discussion revolves around a physics problem involving friction force and work done on an object moving at constant velocity. The original poster attempts to analyze the forces acting on a 15 kg box on a horizontal surface, specifically focusing on the friction force and the work done without an applied force.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of constant velocity on the applied force, questioning whether it must be zero due to zero acceleration. They discuss the relationship between the work done by friction and the work that would be done by an applied force.

Discussion Status

Participants are actively engaging with the problem, raising questions about the nature of the applied force and its relationship to the work done against friction. Some guidance has been offered regarding the conditions of work and energy, but no consensus has been reached on the final interpretation of the problem.

Contextual Notes

There is a noted lack of data precision, as one participant mentions the absence of data to five decimal places, which may affect the calculations being discussed.

kara123
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Homework Statement
How much work is done by pushing a 15 kg box at a constant velocity across a horizontal floor with a coefficient of friction of 0.68 for a distance of 8.0 m?
Relevant Equations
Ff=coefficent of friction x FN
FN=mg
W=fxd
-i had begun by finding the normal force =147 N
-then found the force of friction=99.96 N
-found the work of friction=-799.68 J
after that I am unsure of where to go since I don't have a force applied
 
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What can you say about the applied force?

Hint: you are told velocity is constant.
 
would it be zero since acceleration is zero
 
kara123 said:
would it be zero since acceleration is zero
Acceleration is zero, but there's friction to overcome!
 
so since work of friction equals -799.68 J then work applied must equal 799.68 J or excel it?
 
kara123 said:
so since work of friction equals -799.68 J then work applied must equal 799.68 J or excel it?
It can't exceed it, as then the box would have gained KE.
 
okay so the answer would be 799.68 J of work is done by pushing a 15kg box across a horizontal floor at a constant velocity
 
Yes, except you don't have data to five decimal places.
 
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got it thank you!
 
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