# Kinetic Friction not so easy

1. Sep 20, 2007

### jr4life

1. The problem statement, all variables and given/known data
The block is pushed to the right at a constant velocity. If the coefficient of kinetic friction is 0.5, what is the magnitude of the pushing force?

2. Relevant equations
I dont know...

3. The attempt at a solution

2. Sep 20, 2007

### learningphysics

Did they give the mass of the block? What's your first idea? Give it a shot...

3. Sep 20, 2007

### jr4life

no they didnt give the mass of the block, that is why im so confused my teacher said something about sigma Fx=0 and sigma Fy=0 but i dont know what to do with those numbers, all i know is velocity is constant which makes acceleration 0 which makes force of x and y 0, but after that im lost

4. Sep 20, 2007

### learningphysics

Yes, net force in the x direction is 0. Net force in the y-direction is 0.

What are the forces acting in the y-direction?

What are the forces acting in the x-direction?

5. Sep 20, 2007

### jr4life

In the x direction is the pushing force and force of friction and in the y direction is force of gravity on a box whos mass is not given and the normal force, i think the lack of given numbers is confuing me

6. Sep 20, 2007

### learningphysics

Yeah, just use the variables... let mass = m.

Write the $$\Sigma\vec{F} = ma$$ equation for the x-direction... then for the y-direction... just use the variables...

7. Sep 20, 2007

### jr4life

in x direction it would be (mass)(0) so resultant force would be 0, because acceleration would be 0 due to constant velocity and in y direction (mass)(9.8) and then i dont really know what you could do with that...

8. Sep 20, 2007

### learningphysics

Write the equations using Fpushing, friction, mg, Fnormal... how do they add or subtract?

9. Sep 20, 2007

### jr4life

mg-Fnormal=0

this is where i get confused with x because they are not in equilibrium so they wouldnt equal 0
Fpushing-friction=?? idk

10. Sep 20, 2007

### learningphysics

It equals 0, because acceleration is 0 (constant velocity). Now, you also know that friction = $$\mu * F_{normal}$$

Solve for Fpushing using your two equations, in terms of mass, $$\mu$$ and g.

11. Sep 20, 2007

### jr4life

ok, i understand why it equals zero now but when you use that equation would it be
friction=(0.5)(9.8m) and which two equations are you referring to?

12. Sep 20, 2007

### learningphysics

Yes, that's right...

The two equations I meant were:

Fpushing - friction = 0
Fnormal - mg = 0

you actually used the second equation to get friction = 0.5*9.8*m

So what does Fpushing come out to...

13. Sep 20, 2007

### jr4life

Fpushing-(0.5)(9.8m)=0
Fpushing-4.9m=0
Fpushing=4.9m??

14. Sep 20, 2007

### FedEx

Seems good to me.

15. Sep 20, 2007

### learningphysics

yup. looks good to me too.