Confused why both answers are different. P = F.v

[CeilingFan]

Homework Statement

A fire hose for use in urban areas must be able to shoot a stream of water to a maximum height of 34m . The water leaves the hose at ground level in a circular stream 3.0cm in diameter.
What minimum power is required to create such a stream of water? Every cubic meter of water has a mass of 1.00×103kg.

Homework Equations

Ek = 0.5*m*v^2
Ep = m*g*h
Volume cylinder = A*h
P=F*v (?)

The Attempt at a Solution

I attempted twice, and one of them was wrong (almost double the value), so I was wondering what went wrong in the second attempt.

1st attempt:

First I went to find the speed of the water needed to leave the nozzle.
I used:
Ek(leaving nozzle) = Ep(at 34m)
0.5*m*v^2 = m*g*h

and got
velocity = 25.827 ms^-1

Then I went to find the volume, then the mass of water leaving the nozzle in 1s.
Mass = A*h*1000
pi*0.015*0.015*25.827*1000
= 18.2566kg

With that, I went to find the energy per second required (hence power) to give this mass of water speed V.
Ek = 0.5*18.2566*25.827*25.827 = 6089J

So in Watts, that's the correct answer.

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But the stupid me went and tried another way of solving it (i don't know why i'd ever do that)

So with v = 25.827 ms^-1
I used equation P = F.v for the power needed
P = F.v = (v.dm/dt)v
= v*v*(dm/dt)
So in one second,
Energy to water -> 25.827*25.827*(18.2566) = 12177

Which is the wrong answer, and coincidentally (or not), is twice as large.

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So... I can't figure out what went wrong with the second attempt.

 

[ehild]

Homework Statement

A fire hose for use in urban areas must be able to shoot a stream of water to a maximum height of 34m . The water leaves the hose at ground level in a circular stream 3.0cm in diameter.
What minimum power is required to create such a stream of water? Every cubic meter of water has a mass of 1.00×103kg.

Homework Equations

Ek = 0.5*m*v^2
Ep = m*g*h
Volume cylinder = A*h
P=F*v (?)

The Attempt at a Solution

I attempted twice, and one of them was wrong (almost double the value), so I was wondering what went wrong in the second attempt.

1st attempt:

First I went to find the speed of the water needed to leave the nozzle.
I used:
Ek(leaving nozzle) = Ep(at 34m)
0.5*m*v^2 = m*g*h

and got
velocity = 25.827 ms^-1

Then I went to find the volume, then the mass of water leaving the nozzle in 1s.
Mass = A*h*1000
pi*0.015*0.015*25.827*1000
= 18.2566kg

That is the mass of water leaving the nozzle in 1 second.

With that, I went to find the energy per second required (hence power) to give this mass of water speed V.
Ek = 0.5*18.2566*25.827*25.827 = 6089J

So in Watts, that's the correct answer.

That is the energy /second ΔEk/Δt = 6089J/s =6089 W.

an excellent solution!

----------------------------------------

But the stupid me went and tried another way of solving it (i don't know why i'd ever do that)

So with v = 25.827 ms^-1
I used equation P = F.v for the power needed
P = F.v = (v.dm/dt)v
= v*v*(dm/dt)

I see you applied the impulse law F=Δ(mv)/Δt. Be careful: mass is not created: mass of water is accelerated up to the given speed. P=Fv can be applied when a certain force acts on a certain object with constant mass.


ehild

 

[CeilingFan]

Hi! Thanks for the quick reply!

Do you mean that, only if the force is acting on a constant mass, i.e. F=m*a, can we then use P=F*v?

So if mass is moved/changed and requires dm/dt in the force equation, P=F*v cannot be considered"? (I'm still not too sure about the idea behind P=F*v, so...)

 

[ehild]

P=Fv comes from the definition of work. If the force F acts on an object along a path, the work is equal to the integral $W=\int{\vec F \cdot\vec{dr}}$. $\vec {dr}=\vec v dt$,
so
$W=\int{\vec F \cdot\vec{v}dt}$.
The instantaneous power is $dW/dt = \vec F \cdot\vec{v}$
To apply the formula, you need to know the force. And the force is F=ma. In Classical Physics, the mass is constant and it can not be created from nothing. F=d(mv)/dt has to be used with caution.
In the problem, a certain amount of water has to be accelerated to a certain speed in 1 s. dm/dt is not change of mass, it is the mass flown across the cross-section of the hose in unit time.

ehild