Just need to clarify something (FRICTION)

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To start the motion of a 60kg package on a level floor with a static friction coefficient of 0.56, a horizontal force greater than 329.28N is required. To maintain a constant velocity once the package is in motion, a force of 217.56N is needed, based on the kinetic friction coefficient of 0.37. The discussion confirms that the force required to initiate movement is greater than that needed for constant velocity. The calculations provided are validated by participants in the thread. Overall, the analysis of static and kinetic friction is accurate and well-received.
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QUESTION: If a 60kg package rests on a level floor with coefficients of static and kinetic frictions 0.56 and 0.37 respectively, what horizontal force is required to start the package's motion? how much force is needed to slide the package across the floor at constant velocity?

using the formulas f(static*max*)=mu * N
where mu(static) = 0.56 and N=m*g = 60*9.8 = 588N

so the maximum static friction would be 329.28N... so the force needed to start the package's motion would be any value GREATER than 329.28N

and that means that the kinetic friction is
f(kinetic) = mu * N
where mu(kinetic) = 0.37
so f(kinetic) = 217.56N
and it takes 217.56N of force to slide the package across the floor at constant velocity..

__________________________________

I was just wondering if what i did was correct... and if the force needed to move an object at a constant velocity less than the force needed to move it from rest...

THX>>>>
 
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Your answers look good to me :smile: and welcome to PF.
 
They are? CooL... heheheh

Thanks for the welcome... ^^
 
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