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Homework Help Overview

The original poster seeks to determine the power Superman would need to exert in order to pull a train up an inclined plane. The problem involves concepts from mechanics, including forces, power, and friction, as well as the application of physics equations related to motion and energy.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations of forces acting on the train, including gravitational force, frictional force, and the force applied by the train. There is a focus on understanding the relationship between power, force, and velocity, with some questioning the original poster's understanding of these concepts.

Discussion Status

The discussion is ongoing, with participants providing insights into the calculations and clarifying misunderstandings regarding the equations used. Some guidance has been offered on how to approach the problem, particularly regarding the use of the power equation.

Contextual Notes

There is an assumption of constant velocity for the train, and the original poster expresses uncertainty about the difference between force and power. The discussion also reflects a mix of confidence and hesitation regarding the application of physics principles.

atOnz
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I need to prove how Superman would 'be more powerful than a locomotive.'

http://img221.imageshack.us/img221/5408/diagram8gm.th.jpg

From my diagram I need to find out how much power Superman has to put out to be able to pull that train up the track. I know he has to pull the train 8.5m on an inclined plane of 15 degrees.

-I know the train has a mass of 395 300kg.
-I also know the train is applying a force in the opposite direction.
-Coefficient of kinetic friction of a steel on steel is 0.57
-And finally I know the train is putting out 2,387,200W (3, 200hp) to make that applied force.

So Superman has to overcome the FApplied(train) + FFriction + Fgx to be able to pull the train.
------------------------------
-FN = mgcosx = (395300kg)(9.81m/s^2)(cos15) = 3745757N

-uk = FFriction/FN
FFriction = (0.57)(FN) = 2135081N

-Fgx = mgsinx = (395300kg)(9.81m/s^2)(sin15) = 1003673N
------------------------------

Okay so I found FFriction and Fgx of the train but I still need the FApplied by the train to know how much force superman needs to apply in the opposite direction to move it up the hill.

I know the train is putting out 2,387,200W of power in the opposite direction. But how can I calculate how much force that is? Maybe I misunderstand the difference between force applied and power?

Please help me in anyway possible.
-Jay
 
Last edited by a moderator:
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Assuming constant velocity, you can use the equation P=Fv, where v is velocity. Remember that F and v are vectors.
 
Last edited:
Hootenanny said:
Assuming constant velocity, you can use the equation P=Fv, where v is velocity. Remember that F and v are vectors.

So I researched that this train can get up to at most 56km/h (15.5m/s). I guess the easiest thing to do is assume it's constant because well, to do otherwise might be out of my current level of physics education :P.

2, 387, 200W = F(15.5m/s)
F=154013N

So superman has to overcome Fgx (1003673N) + FFriction (2135081N) + FApplied(train) (154013N) = 3292767N?

Then I guess I can use P = mv again once I find the average velocity a person walks with, to find how much power Superman puts out?
 
atOnz said:
So superman has to overcome Fgx (1003673N) + FFriction (2135081N) + FApplied(train) (154013N) = 3292767N?
Yep, that's some strong biceps there :biggrin:
atOnz said:
Then I guess I can use P = mv again once I find the average velocity a person walks with, to find how much power Superman puts out?
That is the expression for momentum. If you mean to use P = mgv, then no you can't. The force of gravity and the velocity are perpendicular to each other and hence are independent. However, you could work out his power output using P = 3292767v, where v is his walking velocity.
 
Last edited:
Hootenanny said:
Yep, that's some strong biceps there :biggrin:

That is the expression for momentum. If you mean to use P = mgv, then no you can't. The force of gravity and the velocity are perpendicular to each other and hence are independent. However, you could work out his power output using P = 3292767v, where v is his walking velocity.

Oh my, sorry for that! I meant P=Fv. Thank you for your help!
 
atOnz said:
Oh my, sorry for that! I meant P=Fv. Thank you for your help!
My pleasure :smile:
 

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