Simple forces problem no incline

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SUMMARY

The discussion revolves around calculating the horizontal force required to push a 6kg block across a flat table at a constant speed of 0.350 m/s, given a coefficient of kinetic friction of 0.12. The user correctly calculated the weight of the block as -58.8 N and determined the force of friction to be 7.066 N. The final step involves applying the equation Ftotal = ma, where acceleration (a) is zero, confirming that the applied force (F) must equal the force of friction to maintain constant speed.

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Homework Statement


this is simple but I am stuck

in a lab, a 6kg block is pushed across a flat table by a horizontal force F. if the box is moving at a constant speed of 0.350m/s and the coefficient of kinetic friction is .12 what is the magnitude of F


Homework Equations





The Attempt at a Solution



first i drew a picture and figured the object weight would be (m*g) = -58.8 and since it is horizontal the normal force would be balanced with this value. from there i tried to find the force of friction using Force of friction equals normal force*coefficient of kinetic friction and got 7.066N

i need help on where to go from here since the block is not accelerating, and i have a constand speed of .35m/s
 
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Welcome to PF!

Hi HelpMePlz2010! Welcome to PF! :smile:
HelpMePlz2010 said:
in a lab, a 6kg block is pushed across a flat table by a horizontal force F. if the box is moving at a constant speed of 0.350m/s and the coefficient of kinetic friction is .12 what is the magnitude of F

first i drew a picture and figured the object weight would be (m*g) = -58.8 and since it is horizontal the normal force would be balanced with this value. from there i tried to find the force of friction using Force of friction equals normal force*coefficient of kinetic friction and got 7.066N

Yes, that's fine, and you're almost finished.

All you have to do is use Ftotal = ma with a = 0. :wink:

(oh, and the speed is completely irrelevant! :biggrin:)
 
thank you
 

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