Need help finding coefficient of kinetic friction

AI Thread Summary
To find the coefficient of kinetic friction for a 40.0 kg crate pushed with a 150 N force at constant speed, the normal force was calculated as 392 N. Since the crate moves at constant speed, the frictional force equals the applied force of 150 N. The coefficient of kinetic friction is then determined using the formula fK = coefficient of kinetic friction × N, leading to a value of approximately 0.38. This calculation confirms that the frictional force balances the applied force, resulting in no net acceleration. The discussion concludes with the correct coefficient value.
zinniagirl
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Homework Statement


A horizontal force of 150 N is used to push a 40.0 kg packing crate a distance of 6.00 m on a rough horizontal surface. If the crate moves at constant speed, find the coefficient of kinetic friction between the crate and the surface.


Homework Equations


fK= coeff. of kin. friction X N
Fy= -mgy + normal force


The Attempt at a Solution


I started by using Fy= -mgy + N to be 0= -40 (9.8) (cos 0) + N and found N be 392. But now I'm not sure where to go from here to find the coefficient of kinetic friction. Can anyone help?
 
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Constant speed means the resultant force is zero...so what is the value of the frictional force?
 
rock.freak667 said:
Constant speed means the resultant force is zero...so what is the value of the frictional force?

150 N. So the coefficient of kinetic friction would then be .38, correct?
 
Yes that should be correct.
 
Thanks!
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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