Finding Friction Force & Work with No Applied Force

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SUMMARY

The discussion centers on calculating the friction force and work done on a 15 kg box moving at a constant velocity across a horizontal surface. The normal force was determined to be 147 N, and the force of friction was calculated as 99.96 N, resulting in a work of friction equal to -799.68 J. Since the box is moving at constant velocity, the work done by the applied force must equal the work of friction, which is 799.68 J, ensuring no net change in kinetic energy.

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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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