# Im just not getting these force things

1. Oct 3, 2004

### HobieDude16

ok, heres 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? im beginning to hate forces

2. Oct 3, 2004

### ComputerGeek

did you happen to take momentum into account? car crash problems are classic momentum problems.

3. Oct 3, 2004

### HobieDude16

we arent to that part of the book yet, would it still be involved? we havent learned about momentum yet...

4. Oct 3, 2004

any ideas?

5. Oct 3, 2004

### vsage

Hrm I don't see anywhere that you accounted for gravity pushing the car down the hill (mgsin(12)) in part. Well pulling really.

6. Oct 3, 2004

### HobieDude16

one of my classmates whos working with me did this....
MUkmgcos12-mgsin12=m Ax
sooo.... MUkgcos12-gsin12=Ax
then, got Ax, plugged into Vf^2=V0^2+2 Ax D
and solve for Vf (velocity final)....
and that didnt work.... so what would be wrong with that one? or the first one? AH! i hate physics!

7. Oct 3, 2004

### 0aNoMaLi7

i have the same problem and have tried similiar methods...anyone here know what we are doing wrong!?

8. Oct 3, 2004

### HobieDude16

hey oanomali, check your private messages

9. Oct 3, 2004

### vsage

Ok I worked it out and this is what I THINK the solution should look like. Use one of your constant acceleration displacementish formulas:

Our perspective is parallel to the inline of the hill. The resultant force between gravity and friction is as follows:

Fgravity - Ffriction = Fresultant

m*g*sin(12) - u*m*g*cos(12) = m*Aresultant
divide by m.

Note a will be negative. With the equation you used you probably got a positive a. (I had negative/positive switched in original post)

the constant acceleration equation you used from here should work. I think you got a negative sign wrong maybe that was the problem?

Last edited by a moderator: Oct 3, 2004
10. Oct 3, 2004

### HobieDude16

to find t in the d = v1*t + 1/2 *a*t^2 do you complete the square?

11. Oct 3, 2004

### vsage

Don't worry about what I said what I said with d = v1*t + 0.5*a*t^2.. I forgot about the one you used :)

12. Oct 3, 2004

### HobieDude16

ok, using what you said above, i got 1.94 as the acceleration, and plugging that into vf^2=v0^2+2ad with v0 being 16m/s, and d being 25m, we got vf=18.79, which doesnt make sense to me cause thats greater than v0, does that make sense? should the velocity after hitting the breaks be more? it might though, since the friction is minimal, and its downhill.... what do you think?

13. Oct 3, 2004

### vsage

Is this for part b? For part A with my equation I'm getting an accerlation of about -4m/s^2 and for part b about 1.08m/s^2. It seems like you punched a number in wrong or something because using what I said above I get a different answer. That or what I punched into the calculator differs from what I wrote. Let me look.

For part A did you use:

a = 9.8*sin(12) - 0.6*9.8*cos(12)?

Last edited by a moderator: Oct 3, 2004
14. Oct 3, 2004

### 0aNoMaLi7

15. Oct 3, 2004

### HobieDude16

yeah, i think that was part b, but, when i plug it in with 1.08, i still get a speed above the v0, isnt that strange? shouldnt it be lower? 17.6 is what i got for the speed (in part b)... but thats more than v0 which is 16, so what do you think?

16. Oct 3, 2004

### vsage

The speed should be higher for part b than v0. It's right.

17. Oct 3, 2004

### HobieDude16

score! i wasnt sure if i wanted to guess cause i had 1 guess left, but thank you very much, it worked! i wasnt sure with it being faster after sliding, odd....

18. Jan 13, 2005

### atlbraves49

can you guys send me the answers for this problem? or at least how to do it.. Ive been trying to work on it for forever and cant get it. Thanks

Oh and in my problem, the distance is 27m and the v1 was 18m/s instead of 16m/s

thanks

19. Jan 14, 2005

### Leo32

I was just wondering:

How many of you first made a small sketch indicating all forces on the vehicle, before starting to calculate ?

Greetz,
Leo

20. Jan 14, 2005

### jdstokes

You need to find the equation of motion of the car. That means using newton's second law or otherwise to find an expression for the acceleration of the car.
\begin{align*} \Sigma F & = -ma \\ \intertext{the forces acting are friction and the gravitational force,} f_\mathrm{k} - mg\sin\theta & = -ma \end{align*}

The kinetic friction is given by $f_\mathrm{k} = N\mu_\mathrm{k}$ where $N = mg\cos\theta$. Therefore

\begin{align*} \mu_\mathrm{k}mg\cos\theta - mg\sin\theta & = -ma \\ a = g(\sin\theta - \mu_\mathrm{k}\cos\theta) \end{align*}

To find the speed after a distance $s$,

\begin{align*} -v^2 & = -v_0^2 + 2as\\ v & = \sqrt{v_0^2 - 2sg(\sin\theta - \mu_\mathrm{k}\cos\theta)} \end{align*}

Substituting all the known variables gives a speed of 21 m/s for part a and 14 m/s for part b.

The answer to your question depends of the coefficient of kinetic friction.