# Detached train cart

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1. Mar 16, 2015

### Totally

1. The problem statement, all variables and given/known
A train of mass M is moving at uniform velocity v, being pulled by a locomotive with power P. A single cart of mass m gets detached. How far will it go before coming to stop? M, v, P and m are known.

The attempt at a solution
I know I'll need to find coeff. of kinetic friction in order to find the sliding distance. The force on the train before the cart gets detached is
\begin{equation*}
F-F_f=0
\end{equation*}
\begin{equation*}
\dfrac{P}{v}-\mu Mg=0
\end{equation*}
\begin{equation*}
\mu=\dfrac{P}{vMg}
\end{equation*}
Now, after the cart is detached I'm trying to use conservation of energy to find the distance
\begin{equation*}
\dfrac{1}{2}mv^2-\mu mgd=0
\end{equation*}
\begin{equation*}
d=\dfrac{v^2}{2\mu g}
\end{equation*}
Plugging in the $\mu$ expression I get
\begin{equation*}
d=\dfrac{Mv^3}{2\mu P}
\end{equation*}
Using the M, v, P given initially and the kinetic friction that I found from the first part, I get the distance roughly 2.2 times too short when compared to the answer key. I'm sure I did the SI conversions correctly. Any ideas?

2. Mar 16, 2015

### Staff: Mentor

All good except for that µ in the denominator. It should have disappeared.

3. Mar 16, 2015

### Totally

Ah, right, I just substituted it But nevertheless, the answer still does not fit. Should I take it as a mistake in the key? I made an assumption that the velocity is uniform because it said "constant power." I don't think it would be possible to solve it if we say that
\begin{equation*}
F-F_f=ma
\end{equation*}
because there is no way to determine the acceleration from given data.

4. Mar 16, 2015

### BvU

Since we don't have any numbers, it's difficult to check the calculation.
We also don't have the complete problem statement (the 'constant' power appeared out of nowhere). Anyway, once the wagon is detached it doesn't matter .

What you need for your formulas to be applicable is constant friction force.
Since you have no indication that assumption is not allowed, all you can do is make it. (but it's disputable).

Numerically I get 3.2 km for a 400 tonne train moving at 144 km/h when P is 4 MW ($\mu$ = 0.025)

5. Mar 16, 2015

### Totally

Using 600t at 40km/h when P is 10MW, I'm getting 41.14 meters.

6. Mar 16, 2015

### BvU

Must be with the brakes on ! $\mu$ = 0.153
And I do get the same result but find it hard to believe: the deceleration is 1.5 m/s2

But we get the same values with the same expressions.
And the answer key claims 90 m ?

Could you post the full problem statement ?

7. Mar 16, 2015

### Totally

The key demands 600m, what I got before was due leaving the mu in after substitution.

"A train, with a mass of 600t, is moving on a straight, horizontal track at the speed of 40km/h. The locomotive power is constant: 10MW. Suddenly, the last cart with a mass of 60t gets detached. The breaking system fails, and the train of 9 wagons continues onward. The last cart lags behind, gradually getting slower and slower. How far will this cart go until it comes to a full stop?"

8. Mar 16, 2015

### BvU

That's a bit more than a factor of 2.2 !

Can't make sense of the book answer: 10 MJ/s / (40/3.6 m/s) = 900000 N to pull 600 t, so 90000 N to pull 60 t, so a deceleration of 1.5 m/s .
10/9 m/s divided by 1.5 m/s is a standstill in 7.41 s.
SUVAT eqn gives 41.15 m.
kinetic energy 0.5 * 60000 * (10/9)2 = 3700000 J = friction * distance gives 41.15 with a friction of 90000 N.

A distance of 600 m would follow from a friction force of 6173 N. Wouldn't know how to obtain that...

Curious to find out how teacher solves this !

9. Mar 17, 2015

### Totally

Wouldn't be the first problem with a wrong answer. Just had one where, according to the key, a guy could swim faster against a current than in still water.
Glad to know my calculation is solid though, thank you!