ndk123
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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.
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Muttley79
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(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!
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RDKGames
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(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.
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Muttley79
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(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
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ndk123
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Muttley79
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(Original post by ndk123)
....
Did you forget the resistive force?
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