Originally posted by Aaron
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Hi Aaron,
The Joule is a measure of energy. Energy is a measure of Power multiplied by Time.
So, if you use 1W of power for 1 Second, then you have used 1J of Energy.
This structure helps us to understand that the energy dimension already has time embedded in it.
For those reasons, when we say we are using 1J for a given period of time it can be a little confusing because you are essentially adding a time factor to something that already has a time factor in it.
For example - if you say that your 4x4 burns 4 Gallons of fuel for every 60 miles you drive, you can equate that to a specific value of energy (1 Gallon = 1.3 x 10^8 J ). So you could transfer back and forth between miles per gallon (15 MPG) or Miles per 1.3 x 10^8 J (same number - 15 M/1.3 x 10^8 J)). But if you want to relate that to time - you must introduce it from your speed, because it is already embedded in that value. @60mph, it will take 1 Hour. At 30mph it will take 2 hours. At 15 mph it will take 4 hours. But the energy is the same in all 3 of those scenarios, the only thing that is different is how long it took you to use it.
But now let's really talk about the root basis of your question: Acceleration
If I am going to move a mass from one location to another then I have a distance. It will take a specific Force and a Specific amount of time to move from Zero velocity to some constant value that I can apply over that distance.
Back to your 4x4 - You take off from a standing stop and Accelerate to 60mph and then you drive for 58 minutes @ 60mph. We already know how much energy you used to cover that 59miles. But we don't know what it took to get that vehicle up to 60mph from a stand still and we don't know how long that took or how much distance it took. Let's suppose it took a full mile (nice easy acceleration). If your acceleration was constant (linear) , what would it need to be to get to that speed in that distance? Acceleration = Delta Velocity divided by Delta Time. Since we know that we are going 60mph, that is our Delta Velocity (we are supposing here that our acceleration is along a straight line). Now, we have put in a very specific value that forces the time to pop out, the distance of one mile. The distance formula: d = 1/2 ( Vf + Vi ) × t will tell us what the time is in hours where Velocity Final (Vf) = 60 miles per hour and Velocity Initial (Vi) = 0.
This shows us: t in hours = 1 mile / 0.5 x 60 = 1/30 Hours or 120 Seconds.
So our Acceleration must give us 1 mile in 120 Seconds. How do we get the instantaneous acceleration? Using the formula a = (Vf - Vi)/t we get 60MPH / 120 or 0.5MPH per second which would look like 0.5 Miles/hour/s which is the same as saying 8.8 inches per second per second or 8.8"/s². But, in real terms, you would watch your speedometer increase by 1/2mph for each second you are accelerating up to 60MPH.
Now it would be good to know the mass of the vehicle to get to the Joules involved during the acceleration. The Joules are the dimension associated with work and work and work is Force times Distance. Therefore, if we know the mass, we can calculate the force from the acceleration (F=ma) and determine the Work in Joules (W=Fd).
To make things easy I am going to switch to SI units here - meters, kilograms, and seconds(mks).
1 mile = 1609m
Let us say that the 4x4 has a mass of 2268kg (Typical F250 4x4). F=ma: This means our Engine provides a Force of (2268kg * (0.5*1609m)/hours/seconds)Nm or (2268 * 0.2235m/s²)= 506.84Nm of force to accelerate to 60MPH during that two minute period. Of course that is just to move the mass, it does not account for any friction or drag or other typical losses in the drive system. But it is interesting that it represents only 1/150 of the energy contained in a single gallon of gasoline. In other words, we could stop and accelerate like this every 2 miles for 300 miles and only use 1 gallon of fuel if we could apply 100% of the fuel energy to moving the mass.
The work in Joules then is W=Fd where F = 506.84 and d = 1609, so W = 815497.515J
Now for your questions:
If you have 1J of energy in moving an apple of some unknown weight one meter in one second then perhaps we can suppose that it was moved under a consistent acceleration and has some value of velocity when it reaches that one meter mark. Since the definition of one Joule is ((kg · m)/s²)m or (kg · m²) / s² then your apple must weigh 1kg because the only way to get 1J in this case is to multiply (1kg * 1m²) / 1s² anything else in the kg slot will not result in 1J. So our apple has a 1kg mass which on Earth equates to about 2.2Lbs in weight (remember weight and mass are two different things)
So if Wikipedia's Practical Examples says it takes 1J to 'lift' a small apple one meter, then does this mean that they think a small apple weighs 2.2Lbs? You will notice that the article does not specify a time limit. It is similar to our 4x4 using 4 gallons of gas to go 60 miles regardless of how long it took to get there. So there is clearly something different between the energy it takes to move 1kg 1m in 1s and the energy it takes to lift a small apple with no time limit.
The difference is in the 'lift'. Here we have a continuous negative acceleration of 9.8m/s² pulling on that mass. While it takes 1J of energy to move 1kg 1m in 1s horizontally, it would take more to move it vertically against gravity - 9.8 Newtons multiplied by 1 meter, or 9.8J to lift it that one meter. So when they say small apple, they are saying 1kg / 9.81 = 102 grams (3.6 oz) so that the lift is only 1J. Notice that this is time independent. It does not matter how long it takes to lift that 102g, it will always take 1J to lift it 1m against gravity. W = Fd and the Force is 9.8N and the distance is 1m.
So where W = Fd, and F = ma, and a = dV/dt so Work in Joules = d(m(dV/dt)) has time embedded in it. So even when the apple is stationary on a table, there is still a negative acceleration on it of 9.8m/s², or a Force of -9.8 Newtons. This implies that there must be a normal force of +9.8 Newtons (or acceleration of +9.8m/s²)in the opposite direction to balance the two forces to stale mate.
An interesting experiment in this regard would be to lift a ball 1m and drop it inside a quarter-pipe of 1m radius and see how far it rolls on ice horizontally - maybe 9.8m? Does it matter how fast we raise the ball to the starting point?
So using 0.102kg as our apple weight we now see the time involved does not change the energy involved because the energy has the time embedded in it. So the energy is always just the Force multiplied by the distance and in these questions the force is always 1N (F=ma | F=0.102 * 9.8). Applying any other force, to give an alternate acceleration will always result in a non zero-velocity at your destination altitude.
So, supposing then that the destination is always zero velocity at the end of the move:
.75m * (.102kg * 9.8m/s²) = .75J
.50m * (.102kg * 9.8m/s²) = .50J
.25m * (.102kg * 9.8m/s²) = .25J
Now - if you were looking for the Joules involved where a different acceleration were provided to the end of the movement and you wanted to know the resulting velocity as well as the energy involved to get there, then that is a different set of calculations.
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