Im just not getting these force things

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Homework Help Overview

The discussion revolves around a physics problem involving forces, specifically related to a car accident scenario where one car slides into another. The problem includes parameters such as the slope of the hill, distance between the cars, initial speed, and coefficients of kinetic friction for different road conditions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various methods to calculate the final speed of car A upon impact with car B, considering forces such as friction and gravity. Some participants attempt to apply kinematic equations, while others suggest incorporating momentum concepts, despite some indicating they have not yet covered that topic in their studies.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning the validity of their approaches. Some have provided alternative methods and expressed confusion over the results, particularly regarding whether the final speed can exceed the initial speed. There is a mix of attempts to clarify the physics involved and to troubleshoot errors in calculations.

Contextual Notes

Participants mention constraints related to their current curriculum, indicating that they have not yet learned about momentum, which may affect their understanding of the problem. There are also references to specific values and conditions that differ from the original poster's problem setup.

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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
 
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HobieDude16 said:
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

did you happen to take momentum into account? car crash problems are classic momentum problems.
 
we arent to that part of the book yet, would it still be involved? we haven't learned about momentum yet...
 
any ideas?
 
Hrm I don't see anywhere that you accounted for gravity pushing the car down the hill (mgsin(12)) in part. Well pulling really.
 
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!
 
i have the same problem and have tried similar methods...anyone here know what we are doing wrong!?
 
hey oanomali, check your private messages
 
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:
  • #10
to find t in the d = v1*t + 1/2 *a*t^2 do you complete the square?
 
  • #11
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
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 doesn't make sense to me cause that's 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
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:
  • #14
HobieDude16, check your PM :)
 
  • #15
yeah, i think that was part b, but, when i plug it in with 1.08, i still get a speed above the v0, isn't that strange? shouldn't it be lower? 17.6 is what i got for the speed (in part b)... but that's more than v0 which is 16, so what do you think?
 
  • #16
The speed should be higher for part b than v0. It's right.
 
  • #17
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
can you guys send me the answers for this problem? or at least how to do it.. I've been trying to work on it for forever and can't get it. Thanks

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

thanks
 
  • #19
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
atlbraves49 said:
can you guys send me the answers for this problem? or at least how to do it.. I've been trying to work on it for forever and can't get it. Thanks

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

thanks

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.
[itex] \begin{align*}<br /> \Sigma F & = -ma \\<br /> \intertext{the forces acting are friction and the gravitational force,}<br /> f_\mathrm{k} - mg\sin\theta & = -ma<br /> \end{align*}[/itex]

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

[itex] \begin{align*}<br /> \mu_\mathrm{k}mg\cos\theta - mg\sin\theta & = -ma \\<br /> a = g(\sin\theta - \mu_\mathrm{k}\cos\theta) <br /> \end{align*}[/itex]

To find the speed after a distance [itex]s[/itex],

[itex] \begin{align*}<br /> -v^2 & = -v_0^2 + 2as\\<br /> v & = \sqrt{v_0^2 - 2sg(\sin\theta - \mu_\mathrm{k}\cos\theta)}<br /> \end{align*}[/itex]

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.
 

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