## anonymous 3 years ago 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?"

1. anonymous

how do you set up the question?

2. anonymous

I know that m= 25kg

3. anonymous

F=$nuN$

4. anonymous

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. anonymous

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

6. anonymous

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

7. anonymous

oh ok a

8. anonymous

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

9. anonymous

|dw:1352779234119:dw|

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

oh ok

14. anonymous

so how do you get 1.96 m/s^2

15. anonymous

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

16. anonymous

|dw:1352779486046:dw|

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

oh ok

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

$$\Sigma F=ma$$ 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.

25. anonymous

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

26. anonymous

kinda

27. anonymous

im still a bit confused about the Fn= Fg

28. anonymous

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 $$F_x-F_f=0$$, $$F_x=F_f$$.

29. anonymous

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. anonymous

oh ok i got you so far

31. anonymous

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

32. anonymous

im still confused on how to get Fn

33. anonymous

i got Fg= 245

34. anonymous

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

35. anonymous

oh ok

36. anonymous

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

37. anonymous

oh ok so how do you find the friction force

38. anonymous

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

39. anonymous

oh ok

40. anonymous

(.20)(25) ?

41. anonymous

nvm i mean (245)(.2)

42. anonymous

yup. Then after you get friction force, as you see from posts above, the value of friction force is the force exerted.