## YumYum247 one year ago Yelp Me please!!!

1. anonymous

$F_{net}=ma$ $F_{applied}-F_{f}=m\frac{\Delta v}{t}$

2. YumYum247

Normally the push force given to me in N but for some odd reason, it's given in velocity(m/sec). How do i change that m/s into N?!?!?

3. YumYum247

o it gunna look like this....hold up let me draw!!!

4. anonymous

They gave you a CHANGE in velocity (from rest to 1.6 m/s) and a time. Use that to find the acceleration and then it's straight Newton's 2nd

5. YumYum247

|dw:1442450685587:dw|

6. YumYum247

is it gunna be sec^2???????????????????????

7. anonymous

yeah, they don't cancel it's kg-m/s², which is a Newton

8. YumYum247

ooooooooooooooooooooooooooooooooooooooooooooooooo!! i c

9. YumYum247

so 0.72Kg X 0.8m/SEc^2 right

10. YumYum247

11. anonymous

yes. multiply that to get the net force

12. YumYum247

multiply that by wat tho :"(

13. YumYum247

i've already used up all my variables....

14. YumYum247

what's left to multiply tho????????/

15. anonymous

|dw:1442451064756:dw|

16. anonymous

0.72*0.8 is the right side of the equation

17. anonymous

$$F_{applied} - F_f$$ is the left side

18. anonymous

You're solving for $$F_{applied}$$

19. YumYum247

yes i have the mass (0.72kg) and the 0.8m

20. anonymous

|dw:1442451193773:dw|

21. YumYum247

|dw:1442451186851:dw|

22. anonymous

|dw:1442451340214:dw|

23. YumYum247

|dw:1442451278883:dw|

24. anonymous

sort of, you should end up adding 4.51 N

25. YumYum247

oh yah i see ....sorry....yah it should have been Ff and not FN

26. YumYum247

Alright Thanks.....

27. anonymous

you're welcome