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Mechanics help

The frictional resistance to the motion of a train is k time the combined weight of the engine and its coach. An engine of mass M works at a constant power P and attains a max speed U when pulling a coach of mass M up a hill of inclination x to the horizontal. Down the same hill with a coach of mass M/2, the engine can attain a maximum speed 2U. Show that P is 12kMgU/5.

The engine pulls a coach of mass 3M/2 on level ground. Find the accelaration of the train when the speed U/5 and show that the max speed is (24/25)U.
Original post by ndk123
The frictional resistance to the motion of a train is k time the combined weight of the engine and its coach. An engine of mass M works at a constant power P and attains a max speed U when pulling a coach of mass M up a hill of inclination x to the horizontal. Down the same hill with a coach of mass M/2, the engine can attain a maximum speed 2U. Show that P is 12kMgU/5.

The engine pulls a coach of mass 3M/2 on level ground. Find the accelaration of the train when the speed U/5 and show that the max speed is (24/25)U.


Post what you have got so far and we'll help with the next step.
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ndk123
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Original post by Muttley79
Post what you have got so far and we'll help with the next step.


I know i need to use p=force*velocity
But thats it! Dont rly know where to start!
Original post by ndk123
I know i need to use p=force*velocity
But thats it! Dont rly know where to start!


A good start is to draw a diagram and make sure you know what's going on. Model the situation.
Original post by ndk123
I know i need to use p=force*velocity
But thats it! Dont rly know where to start!


So look at the forces as the train goes up the hill - at max speed acceleration is zero so the forces will be in equilibrium.

Force up the slope = Driving force of train = P/u

Force down the slope/resisting the motion = .....
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ndk123
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Original post by Muttley79
So look at the forces as the train goes up the hill - at max speed acceleration is zero so the forces will be in equilibrium.

Force up the slope = Driving force of train = P/u

Force down the slope/resisting the motion = .....


Thank you! Managed to do the first bit but i keep getting the wrong answer for acceleration, posted a photo of working below
Reply 6
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ndk123
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5B30E1A6-B5CB-4B18-9CB6-AD59CE1E3D96.jpg.jpeg
Original post by ndk123
....


Did you forget the resistive force?

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