Can both of these equations be right?

[Dorothy Weglend]

In solving my physics problems, I often end up with two equations for the same thing, they both seem right to me, and yet, it seems, they both can't be right.

I thought I would ask about this here.

I'm solving a problem which involves a box in the front of a railroad car. There is initially no motion, then the car starts to accelerate. I am asked to find various things about the box at the point when it hits the back of the car.

I started by trying to find the acceleration of the box. I get that from the kinetic friction:

fk = ma, or umg = ma, so a = ug, where u is the coefficient of kinetic friction.

With respect to the railroad car, we have

a - A = ug - A, where A is the accleration of the railroad car. So far, so good, I think. Now this is where my problems start.

If the distance the box travels backwards in the railroad car is l, then the time it takes for the box to contact the back of the railroad car is the same as it takes the car to advance l, so

l = At^2/2, and t = sqrt(2l/A).

So the velocity of the box, relative to the car, should be

v = (ug - A)sqrt(2l/A)

Of course, you can also use v^2 = 2as to do the same thing. I tried that:

v^2 = 2l(ug - A), so v = sqrt(2l(ug-A))

As far as I can tell, these aren't equivalent expressions, mathematically, for the same quantities, but I think my reasoning is sound in both cases.

This happens to me ALL the time. Can someone help me figure out which one of these is right?

Thanks,
Dorothy

 

[Doc Al]

With respect to the railroad car, we have

a - A = ug - A, where A is the accleration of the railroad car. So far, so good, I think. Now this is where my problems start.

Yes. So far, so good!

If the distance the box travels backwards in the railroad car is l, then the time it takes for the box to contact the back of the railroad car is the same as it takes the car to advance l,

What makes you think that? According to that line of reasoning, with respect to the tracks, the box doesn't move! (If the car moves back a distance l with respect to the train, but the train moves forward a distance l in the same time...)

so

l = At^2/2, and t = sqrt(2l/A).

This is the time it takes the train to move a distance l with respect to the tracks.

So the velocity of the box, relative to the car, should be

v = (ug - A)sqrt(2l/A)

This is the relative speed after the train moves a distance l--not what you want.

Of course, you can also use v^2 = 2as to do the same thing. I tried that:

v^2 = 2l(ug - A), so v = sqrt(2l(ug-A))

This time, since you are using the acceleration of the box with respect to car, you are finding the relative speed after the box moves a distance l within the car. This is what you want.

As far as I can tell, these aren't equivalent expressions, mathematically, for the same quantities, but I think my reasoning is sound in both cases.

No, they are not equivalent. Each one sneaks in a different assumption. If you want to know what's happening at the moment the box reaches the back of the car, then use the acceleration (and distance) with respect to the car!

 

[Dorothy Weglend]

What makes you think that? According to that line of reasoning, with respect to the tracks, the box doesn't move! (If the car moves back a distance l with respect to the train, but the train moves forward a distance l in the same time...)

Well, you are right. I was thinking that the box didn't move. But of course, there is friction, which must move the box, so the train could move considerably farther than l to get the box all the way to the back of the train. Sheesh...

Thank you, Doc Al. You have helped me so many times, you are just great. I really appreciate it. I hope someday I can help people just like you do.

Dorothy