How far did the train travel before stopping after detaching a compartment?

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SUMMARY

The discussion centers on calculating the distance a train compartment travels after detaching from a train moving up a 4-degree incline at a speed of 5 m/s. The initial kinetic energy (KE) of the compartment is converted into potential energy (PE) as it ascends the incline. The correct approach involves using the equation 0.5 * mv² = mgh to find the height reached, followed by trigonometric calculations to determine the distance along the incline, resulting in a distance of approximately 18 meters.

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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of motion on inclined planes and energy conservation principles.

mrhobbes
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A train moving up an incline at 4 degrees has it's last train compartment detached. The train was moving at a speed of 5 m/s when the compartment breaks free. How far did the train get before the speed momentarily reaches zero?

No other information is given, but using all that was provided, I got a ridiculously small answer of 0.09 m. I got this by splitting up my speed into x and y components and then using the Vo(y) in the equation d= vf² - vo² / 2a to get my vertical displacement. I then used distance to get my inclined displacement by using sin 5 = y / r.

I checked the answer and it turns out that it was wrong. Could anyone give me any hints on how to solve the problem?

Thanks
 
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Welcome to PF.
The question doesn't make much sense as written. Perhaps it means a train CAR breaks free from the back of the train. Without engine. How far does the CAR go up the hill?

It's initial speed is 5 m/s. Since it is going up the hill, its KE will soon be converted into PE. You could begin with
initial KE = final PE
.5*mv² = mgh
and quickly find the height reached from the initial position. You'll have to convert that into a (larger) distance along the ramp with a bit of trigonometry. I'm getting a little more than 18 m.
 
Thank you so much. You were right.
 
Most welcome! Good luck with the next one.
 

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