# Newton's Laws Help FAST! THANKS!

1. Dec 14, 2004

### iamgod21

hey all, ok
so i have this physics problem which i dont understand completely so if i could get some help that'd be great! so here goes:
A 1500kg car is being pulled up a ramp at an angle of 18 degrees. Its being pulled by a tension of T at 27 degrees from the hill. *or 45 degrees* . I need to know the acceleration of the car if the max tension can be 8000N. Also, i need to know the distance it would travel from stop in 5 seconds. Thanks! *Assume frictionless surface*

Heres what i did:
---i set up my FBD w/ my tension and my Fn, and my mg. Set my coordinate systems to up the hill and up towards the sky. So my equations are as follows.
EFx = 8000cos27 - mhsin18=max
EFy = FN + 8000sin27 - mgcos18 = may=0
so i got acceleration was equal to 1.7236666 m/s*s
then i used D=vit + 1/2 a t *t and got the distance to be 21.55 is this correct?
Thanks a billion!

2. Dec 14, 2004

### Staff: Mentor

Looks good to me. Note that there was no need for the Fy equation, but no harm done. Also, beware of quoting too many significant digits in your answer.

3. Dec 14, 2004

### iamgod21

thanks a ton!

4. Dec 14, 2004

### iamgod21

k so i just realized there was more to it so i did that and heres what it is:
now we have to assume theres kinetic friction on the hill at 18 degrees. the coefficient of kinetic friction is .15 and now we need the new acceleration and the new distance
so heres what i did:
i set up new equations:
EFx = 8000cos27 - mgsin18-Fk = max
EFy = FN + 8000sin27-mgcos18=may=0
so FN = 10348.61 N
so i solved Fk and plugged that into my Fx equations and plugged in 1500 for the mass and got that the acceleration was .689
then i plugged that into the same equation as before and go tthat the distance was 8.61 meters...again did i do this correctly???

5. Dec 14, 2004

### Staff: Mentor

It's all good.

6. Dec 18, 2004

### iamgod21

Thanks!!!! I got it correct!