Simple forces problem no incline

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A 6kg block is pushed across a flat table at a constant speed of 0.350 m/s, with a coefficient of kinetic friction of 0.12. The weight of the block is calculated as 58.8 N, and the normal force balances this value. The force of friction is determined to be 7.066 N. Since the block is not accelerating, the total force (F) applied must equal the force of friction. The speed of the block does not affect the calculation, as it is constant.
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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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