The Law of Conservation of Energy states that energy can neither be created nor destroyed, but it does have the ability to change forms. Take for example an object of mass ( m ) that has been raised to some arbitrary height ( h ). The work ( W ) done on the object is a product of its weight ( Fw ) times the distance it has been raised above the earth’s surface:
W = Fd
W = ( Fw )( h )
Fw = mg
W = mgh → Stored as potential energy
The work done in raising the object is stored in the form of potential energy ( PE ). This energy status will remain the same until the object is moved to either a higher or lower elevation. If the object is allowed to fall back to its original position, the potential energy will be converted into the energy of motion called kinetic energy ( KE ):
KE = ½ mv2
We now begin with a before-and-after scenario that relates the initial ( PE1 ) and ( KE1 ) of the system to the final ( PE2 ) and ( KE2 ) within it:
KE1 + PE1 = KE2 + PE2
½ m1v12 + m1gh1 = ½ m2v22 + m2gh2
Fortunately, the equation can be somewhat simplified due to three physical realities present in the system: ( 1 ) The mass of the falling object doesn’t change, so the mass terms ( m1 and m2 ) cancel out of the equation; ( 2 ) The initial velocity ( v1 ) of the object is zero. For this reason, the kinetic energy term on the left-hand side of the equation is equal to zero; ( 3 ) The final height ( h2 ) of the object is zero upon returning to its origin, so the potential energy term on the right-hand side of the equation can be eliminated as well:
v1 = 0 m/s
h2 = 0 m
m1 = m2
0 J + gh1 = ½ v22 + 0 J
Solving the final velocity of the object has now been reduced to a relatively simple task:
v22 = 2gh1
v2 = √ 2gh1
Let’s now examine a system in which a fluid-filled container has its contents empty onto the ground at a constant velocity through a spigot:
Rather than a single particle falling from a high-to-low position, we have an entire fluid doing so. This system of particles contains both ( PE ) and ( KE ) as well. The potential energy is due to the fact that the fluid’s particles are elevated within a container with a height ( h ). Additionally, the stream of liquid moving through the spigot does so with a constant velocity ( v ) and kinetic energy. It would be extremely difficult ( if possible ) to account for the travel made by each atom or molecule within the fluid at hand. For this reason, we must relate the density ( ρ ) of the fluid to the energy ( E ) per unit volume ( V ) of the total quantity of fluid within the container. This is accomplished by converting the equation for pressure from units of force ( F ) per unit area ( A ) to units of ( E ) divided by ( V ):
P = ( F / A )
W = Fd
V = Ad
P = ( F / A )( d / d ) = ( Fd / Ad ) = ( Energy / Volume )
What this shows is that each unit volume of fluid passing through the spigot has the capacity to perform work. We must now develop terms that appropriately express how the energy per unit volume of the fluid mass is conserved in the before and after scenario for the system. Let’s begin by developing an appropriate term for ( KE ) per unit volume ( V ):
P = ( KE / V ) = ( ½ mv2 / V )
The fluid’s density is its mass per unit volume:
ρ = ( m / V )
For this reason, the equation above can be substituted with the density variable:
P = ( ½ )( m / V )( v2 )
P = ½ ρv2
A similar process allows us to develop an expression for ( PE ) per unit volume ( V ):
P = ( PE / V ) = ( mgh / V )
ρ = ( m / V )
( m / V )( g )( h )
P = ρgh
We are almost there 🙂. The last physical reality to consider is that regarding the initial pressure ( P1 ) and final pressure ( P2 ) that imposes forces upon the fluid. We do so by adding each of these terms to the equation that relates the ( PE ) and ( KE ) per unit volume of the initial and final system:
P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + ρgh2
The top of the container and spigot are both exposed to the atmosphere, so P1 and P2 are equal to atmospheric pressure. This gives us the good fortune of eliminating P1 and P2 from our equation:
P1 = P2 = Atmospheric Pressure
½ ρv12 + ρgh1 = ½ ρv22 + ρgh2
Furthermore, we are once again able to eliminate terms due to zero values for ( h2 ) and ( v1 ):
Note: The velocity of fluid at the top of the container is negligible compared to that flowing through the spigot. For this reason, the movement of atoms and/or molecules in this region of the container is approximated as being zero:
h2 = 0 m
v1 = 0 m/s
v2 = ???
0 J + ρgh1 = ½ ρv22 + 0 J
The densities of the fluid cancel out as well. This enables us to solve for ( v2 ) with relative ease:
gh1 = ½ v22
v22 = 2gh1
v2 = √ 2gh1
Wow!!!
This is the same expression that models the velocity of a single object falling from the container’s height downward to the earth below. The key point is that both systems contain energy that must be conserved, but that conservation is expressed differently from one system to another.
Finally, notice how the masses canceled out in each expression. If an object of ( 1 kg ) falls through space, it is acted upon by a force. If a two kilogram ( 2 kg ) object falls through space, it will be acted upon by twice the amount of force:
Fw = mg
Fw1 = ( 1 kg )( g )
Fw2 = ( 2 kg )( g ) = 2 Fw1
The second object is acted upon by twice the amount of force, but each object accelerates at ( g ) near the surface of the earth. Mass ( m ) is a measure of inertia ( I ). The more inertia an object possesses, the more difficult it is to put it into motion, alter its motion, or bring it to rest. For this reason, an energetically valid comparison can be made between a system in which a single particle’s ( PE ) and ( KE ) change as it falls to the earth below and that of a fluid within an equal height container moving towards the earth through a spigot.
Q: A 0.2 m container is full of a fluid of unknown density ( ρ ). A spigot at the bottom of the container is opened to allow fluid to flow at an unknown velocity ( v ) onto the ground. With what velocity will the fluid flow through the spigot?
A: Both the spigot and top of the container are open to the atmosphere. For this reason, the pressure imparted at each surface by the atmosphere is the same. This allows us to simplify the equation that relates energy per unit volume ( P1 and P2 ) of the initial and final systems at hand:
P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + ρgh2
P1 = P2 = Atmospheric Pressure
½ ρv12 + ρgh1 = ½ ρv22 + ρgh2
0 J + ρgh1 = ½ ρv22 + 0 J
gh1 = ½ v22
v22 = 2gh1
v2 = √ 2gh1
v2 = √ [ ( 2g )( h1 ) ]
v2 = √ [ ( 2g )( 0.2 m ) ]
v2 = √ [ ( 2 )( 9.86 m/s2 )( 0.2 m ) ]
v2 = √ 3.944 m2/s2
v2 = 1.99 m/s
Bonus Q: An object of unknown mass falls from a height of 0.2 m. What is the value of its final velocity ( v ) just before it strikes the earth’s surface?
A: ½ m1v12 + m1gh1 = ½ m2v22 + m2gh2
0 J + gh1 = ½ v22 + 0 J
gh1 = ½ v22
v2 = √ 2gh1
v2 = 1.99 m/s
RENAISSANCE PHYSICS 