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

• CeilingFan
In summary, the fire hose needs a minimum of 6089 watts of power to create a stream of water at a height of 34 meters.
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.

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

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.

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

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

CeilingFan said:

## 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.

CeilingFan said:
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!

----------------------------------------
CeilingFan said:
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

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...)

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

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!

## What is the equation P = F.v used for?

The equation P = F.v is used to calculate power, which is the rate at which work is done or the rate at which energy is transferred.

## What do the variables F and v represent in the equation P = F.v?

In the equation P = F.v, F represents force and v represents velocity. Force is a measure of the push or pull on an object, while velocity is the speed and direction at which an object is moving.

## Why are there two different answers when using the equation P = F.v?

There may be two different answers when using the equation P = F.v because it depends on the units used for force and velocity. If different units are used, the resulting power will also be in different units. Additionally, there may be slight variations in the measurements of force and velocity, leading to slightly different answers.

## How can I ensure that I get an accurate answer when using the equation P = F.v?

To get an accurate answer when using the equation P = F.v, it is important to ensure that the units of force and velocity are consistent. If necessary, convert the units to be in the same system (e.g. both in SI units or both in imperial units). It is also important to use precise and accurate measurements for force and velocity.

## Can the equation P = F.v be used for all types of motion?

The equation P = F.v can be used for all types of motion as long as the motion is in a straight line and the force and velocity are in the same direction. If the motion is not in a straight line or the direction of force and velocity are not parallel, then this equation may not be applicable.

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