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

 

[HalfLostAndSearching]

Hello,

Thank you for sharing your thought process and attempts at solving this problem. It is important to analyze and understand why there are discrepancies in your answers.

First, let's review the equation P = F.v. This equation represents the power (P) required to move an object with a force (F) at a certain velocity (v). In this case, the object is the water and the force is the pressure from the hose, which is exerted on the water to move it at a certain velocity. However, this equation assumes that all of the force is used to accelerate the water, which is not the case in this scenario.

In your second attempt, you used the equation P = F.v to calculate the power needed to give the water a velocity of 25.827 m/s. However, this does not take into account the force required to lift the water to a height of 34m. In other words, the force used to lift the water is not the same as the force used to accelerate it.

In your first attempt, you correctly used the equations for kinetic and potential energy to find the velocity of the water and the mass of water needed to reach a height of 34m. This approach takes into account both the force required to accelerate the water and the force required to lift it to a certain height.

Therefore, the discrepancy in your answers is due to the fact that in your second attempt, you only considered the force required to accelerate the water, while in your first attempt, you considered both the force required to accelerate and lift the water.

I hope this explanation helps to clarify the difference in your answers. It is always important to carefully consider the physical principles involved in a problem and choose the appropriate equations to use. Keep up the good work in your scientific pursuits!