# Work concept question

GuN

## Homework Statement

Three different cars (a red one, a blue one and a green one) are driving with an identical velocity. Suddenly they all slam on the brakes and skid to a stop. The red car is the most massive and the blue one is the least massive. Each of the three cars has identical tires. Which car travels the longest distance while stopping?

## Homework Equations

W=F*dcos(theta), F=m*a

## The Attempt at a Solution

My reasoning is that since F=ma, W=m*a*d*cos(theta), so rearranging for d, I get

d=W/(ma*cos(theta)

So greater the mass, the lower the distance. So I guess since m and d are inversely proportional to one another, the blue car (the least massive) travels the longest distance.

Sorry I just want to double check, I managed to get a practice midterm, but with no solutions so I just want to see if my method is correct.

nasu
Why would you assume that the work is the same for all three cars?
What work is that? Of what force?

Velocityed
yea, the work done is different so we can't apply those formulas,
I think they are going to stop at the same time,

barryj
Assuming they all have the same coefficient of friction, do a free body diagram and find what the acceleration is for the three cases.

1 person
GuN
Yeah, it was a conservation of momentum question. The professor slipped it into the work section to trick people.

I asked a TA and he said redo the question with the concept that momentum initial=momentum final, so I assumed the least massive car would have the greatest velocity(because mivi=mfvf) . Since distance and velocity are directly proportional (d=v/t), that means the fastest car travelled the most distance (even though initially they have the same speed, the final speed is different).

Anyways, thanks guys.

barryj
GuN, your above comment is confusing and incorrect. Do what I suggested in post 4. I assume you know nhow to compute the frictional force on each car while skidding to a stop. Then find the acceleration. I think you will find that all cars will have the same acceleration and given the same initial velocity, wellll you go from here. I can't give you the answer.

nasu
Yeah, it was a conservation of momentum question. The professor slipped it into the work section to trick people.

I asked a TA and he said redo the question with the concept that momentum initial=momentum final, so I assumed the least massive car would have the greatest velocity(because mivi=mfvf) . Since distance and velocity are directly proportional (d=v/t), that means the fastest car travelled the most distance (even though initially they have the same speed, the final speed is different).

Anyways, thanks guys.

The momentum of the car is not conserved. Friction is an external force (to the cars) which cannot be neglected. The final momentum of each car is zero. The initial momentum is not zero. Isn't obvious that this NOT a conservation of momentum question?

GuN
That's what the TA told me , and I managed to find the solutions, and it got me the right answer.

I'll ask the professor though. Also, for most of these problems, (at this stage) in introductory physics we don't really acknowledge friction, so I'm not really sure how to incorporate that.---I guess my course is too simplistic to account for that (it's just algebra based physics, and we didn't really get that far yet).

Ha, sorry. I'm going to try ask another TA, or even the professor if I get a chance.

barryj
If there is no friction, then slamming on the brakes will have no effect. I think you should get another TA.

1 person
GuN
Yeah, I found another TA, they're all going to stop at the same time. I think that's what you were hinting at (same velocity, same acceleration)?

Most TAs run the help sessions at different times, and that's the only help we get, so I'm kind of stuck with what TA I'm given because they're only open for a few hours a day.

I'm guessing the solutions were wrong. I got it from a friend, so I'm guessing it's unofficial (a grad student did the solutions for some ). I'm guesing it was a poorly done job.

Thanks guys.

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