- #1
HobieDude16
- 70
- 0
ok, here's another force problem, that i thought i had right, but obviously not, once again...
You testify as an "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill (Fig. 6-27). You find that the slope of the hill is = 12.0°, that the cars were separated by distance d = 25.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 16.0 m/s.
(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
m/s
(b) What was the speed if the coefficient of kinetic friction was 0.10 (road surface covered with wet leaves)?
m/s
what i did:
i found the acceleration by doing a=-fk/m, (fk=MUk(Fn)) (Fn=mgcos12)
so that is a=-((.60)mgcos12)/m)
then i used v^2=v0^2-2a(distance)
so then that would be v^2=16^2-2(-.6gcos12)(25)
but that was wrong too, whatd i do here? I am beginning to hate forces
You testify as an "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill (Fig. 6-27). You find that the slope of the hill is = 12.0°, that the cars were separated by distance d = 25.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 16.0 m/s.
(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
m/s
(b) What was the speed if the coefficient of kinetic friction was 0.10 (road surface covered with wet leaves)?
m/s
what i did:
i found the acceleration by doing a=-fk/m, (fk=MUk(Fn)) (Fn=mgcos12)
so that is a=-((.60)mgcos12)/m)
then i used v^2=v0^2-2a(distance)
so then that would be v^2=16^2-2(-.6gcos12)(25)
but that was wrong too, whatd i do here? I am beginning to hate forces