## abannavong Group Title Physics help with friction! the problem is "You push a 25kg wooden box across a wooden floor at a constant velocity of 1.0 m/s. If the coefficient of friction is .20, how much force do you exert on the box?" one year ago one year ago

1. abannavong Group Title

how do you set up the question?

2. abannavong Group Title

I know that m= 25kg

3. Sea15era Group Title

F=$nuN$

First, you draw the free body diagram. And note that the question tells you: there is no acceleration(constant velocity) Then you apply Newton's second law to it.

5. abannavong Group Title

ok so newtons second law you would use f=ma?

yes. $$\Sigma F = m\Sigma a$$

7. abannavong Group Title

oh ok a

8. abannavong Group Title

so we know that m=25 kg and what is the a?

9. Sea15era Group Title

|dw:1352779234119:dw|

10. Sea15era Group Title

first your normal force equals your mass times gravity (245N) and then plug them into your equation to find a. a=.20x245/25

well, since it is constant velocity, a=0.

12. Sea15era Group Title

for your a you should get 1.96 m/s^2

13. abannavong Group Title

oh ok

14. abannavong Group Title

so how do you get 1.96 m/s^2

Hmm... @Sea15era why is there acceleration from the total force?

16. Sea15era Group Title

|dw:1352779486046:dw|

but the box is moving with constant velocity, i.e. no acceleration, right?

18. Sea15era Group Title

thats just a velocity acceleration is in m/s^2 not m/s i believe.

19. Sea15era Group Title

F=25x1.96 F=49N should be the answer if i did my equations and sum of forces statement wrong.

20. abannavong Group Title

oh ok

21. Sea15era Group Title

I hope this helps you , good luck and if by chance it's wrong I'm sorry!

Yes. The acceleration is not given, but since you're pushing it in constant velocity, there should not be any acceleration.

23. abannavong Group Title

so how do you find the force that is being exerted?

$$\Sigma F=ma$$ Start with this picture. |dw:1352781084312:dw| As shown, the forces are F_x and F_f, of different directions. Applying the second law, $$F_x - F_f =ma=0$$ as it is constant velocity.

Now, sea told you $$F_f=\mu F_N$$, where $$F_N=F_g$$ in this case.|dw:1352781249983:dw| Do you follow?

26. abannavong Group Title

kinda

27. abannavong Group Title

im still a bit confused about the Fn= Fg

Now, as $$F_g = m g$$, where m=25kg and g=9.8, so you get $$F_g$$ Now after you get $$F_g$$ you can get $$F_N$$ After you get $$F_N$$, You get $$F_f$$ Then since $$F_x-F_f=0$$, $$F_x=F_f$$.

On the y-axis, v=0,(it doesn't move vertically), so a=0 too. Since $$\Sigma F_y=F_N-F_g= 0$$,it follows that $$F_N=F_g$$

30. abannavong Group Title

oh ok i got you so far

Step by step, have you got the value of $$F_N$$?

32. abannavong Group Title

im still confused on how to get Fn

33. abannavong Group Title

i got Fg= 245

yup, and that's the value of Fn. Fn is the normal force of the ground acting on the box.

35. abannavong Group Title

oh ok

yeah, so next you will get the value of friction force...

37. abannavong Group Title

oh ok so how do you find the friction force

$$F_f = \mu F_N$$ where $$\mu$$ is the coefficient of kinetic friction.

39. abannavong Group Title

oh ok

40. abannavong Group Title

(.20)(25) ?

41. abannavong Group Title

nvm i mean (245)(.2)