# Homework Help: Increasing a Train's Power

1. Aug 5, 2012

### Tiba

1. The problem statement, all variables and given/known data

A train is traveling on a horizontal and straight track, with a constant speed of 20m/s, when it begins to rain.
The rain is rigorously vertical and the mass of the water that falls over the train and after that runs vertically by it sides is 100kg per second.
Consider the resistance forces (air resistance and the friction in the non-movable wheels) to be constant.
For the speed of the train to keep constant after the rain start, the power developed by the train should suffer an increase of...?

2. Relevant equations

P=F.V

3. The attempt at a solution

I really have no clue... I could only think that the mass of the train will suffer an increase of 100kg, but I don't how to apply this mass increase to the formulas of power, P=T/t and P=FV

2. Aug 5, 2012

### CWatters

Assuming the rain sticks to the train and is accelerated horizontally then..

Every second the train has to accelerate an additional mass of 100kg from 0 to 20m/s.

The KE it needs to give that extra mass is given by

Energy = 0.5*Mass*Velocity2

Power = Energy/Time

3. Aug 5, 2012

### Tiba

Thanks,

So the KE = 20kJ and the Power = 20kW.
However, the answer on the textbook is 40kW.
What am I doing wrong?

4. Aug 5, 2012

### Tiba

I just thought... The force necessary to accelerate the mass of water is F=m.a, and a=V/T (V = 20,0m/s and t=1,0s). So F = 2,0.10³
Energy = Force*Space (Space = 20,0m, because V=20,0m/s and t=1,0s)
Energy = 2,0.10³.2,0.10 = 4,0 . 10^4
and finally
Power = 40,0kW

What mistake I did in the other method using the KE?

5. Aug 5, 2012

### Staff: Mentor

Space is not [strike]a*t[/strike] v*t when the velocity is not constant; Here the motion is accelerated. You need to use another expression for the distance covered.

Last edited: Aug 5, 2012
6. Aug 5, 2012

### bigerst

i think i got 40 kw as well
the train experiences a force, F=dp/dt =d(mv)/dt = v(dm/dt) = v * 100kg=2,000N
P=F*v = 2,000 N *20m/s = 40 kW
well what was wrong with the energy situation?
P* dt = dE = 1/2 dm v^2 = 100kg/s * dt *1/2 *v^2
p = 20kw
now there appears to be an inconsistency, any explanations?

7. Aug 5, 2012

### bigerst

it appears the energy neeeded to acclerate 100kg to 20m/s is 20 kW but the engine needs to supply more than that energy to keep the thing running?

8. Aug 5, 2012

### bigerst

OMG i just realized this situation constitutes an inelastic collision in which case energy is NOT conserved!!!! that resolves the discrepance!!

9. Aug 5, 2012

### Tiba

But I took the velocity of the train as a base, which is constant, not the water's. After all, the power in question is the train's power.

10. Aug 5, 2012

### lewando

"The rain is rigorously vertical and the mass of the water that falls over the train and after that runs vertically by it sides is 100kg per second."

This implies that the train is not accumulating mass (other than an insignificant mass associated with it being wet) and that the rain's final velocity is equal to that of the train.

In a one second interval, getting 100kg to go from 0m/s to 20m/s requires 1/2mV2 (20kJ) of energy (expending energy at the rate of 20kJ per second gives you 20kW of power).

But this energy is not free and comes at the expense of the train slowing down a bit.

How much energy in the one second interval would be required for the train to restore its speed back to 20m/s?

Last edited: Aug 5, 2012
11. Aug 5, 2012

### Staff: Mentor

But if you're using F=m*a and a = V/T, the a must apply to the thing being accelerated, while the V is final velocity.

12. Aug 6, 2012

### Villyer

This may be a coincidence, but I feel like it is interesting to point out.

P=Fv
P=m/t*v*v
m/t = 100, and v*v = 20*20 = 400
So P = 40kW, which is the answer in his book.