How Do You Convert Newtons to Watts for a Maglev Train Acceleration?

[Jon Winter]

Homework Statement

Imagine a maglev train car on a track. The train car weighs 1000 kilograms and travels at 10 meters per second. Without any external force, the train car slows down to 5 meters per second in 10 seconds. How much force does it take to return the train car to traveling at 10 meters per second? Convert that force to watts.

Homework Equations

F=MA
A = (vf -vi)/t
W = (N*m)/s

The Attempt at a Solution

· F = M*A

· F = (1,000 kg)*((10 m/s – 5 m/s)/10 s)

· F = 5,000 kgm/s^2

· F = 5,000 Newtons

· Watt = (N*m)/s

· W = (5,000N * 5m)/10s

· W = 2,500
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[lewando]

Without any external force, the train car slows down to 5 meters per second in 10 seconds

Okay, so there must be some kind of "internal" force that makes this happen. Just an observation-- this information is irrelevant to the problem.

How much force does it take to return the train car to traveling at 10 meters per second?

Are you sure you have transcribed the original question completely? Clearly an element of time is needed to determine this.

 

[haruspex]

F = (1,000 kg)*((10 m/s – 5 m/s)/10 s)

You seem to have assumed the force is applied for 10s. That is not mentioned in your problem statement. Did you leave it out?

W = (5,000N * 5m)/10s

Where does the distance of 5m come from?

 

[Orodruin]

Someone needs to point out that force and power are different physical quantities and that you cannot convert one to the other. What you can do is compute the work a particular force does or the power it is putting into a system. As others have noted, the problem statement is missing information to do that.

Furthermore, a system on which no external force acts does not accelerate by definition.

 

[lewando]

Furthermore, a system on which no external force acts does not accelerate by definition.

Apologies for my poor choice of words: "internal force".

 

[J Hann]

What seems to be missing is the implication that the train speeds back to 10 m/s in 10 sec.
Work (J) = Change in KE = Force * Distance
Power (Watts) = Work / time