Mechanical energy.

[Moha99]

Hallo everyone.

If i had a motor for example that is able to take 110J(theatrically use of simple number to understand easily) of electricity and convert it to mechanical energy. Due to countless energy losses that motor was able to only take let say... 100J form that input energy and the 10J lost.

Now here comes the theocratical example: The motor will be able to do mechanical work lets say there are two cubes that need to be lifted. I connected the motor's shaft with some mechanical gear and lifted the cubes over a 1Meter distance upward.

Now if I lifted 1 cube(without including its weight lets make it a general thing not an exact measurement just an image) 1m up it will receive that 100J of mechanical energy(without energy loss lets theoretically say that).

Now if there wer 2 cubes lifted up 1 meter high would the 100J be divided to each? one cube would take 50J and the other the same? or both would take 100J? (I understand the mass has an important role but lets say the motor can handle the cubes weight 100J is enough for that.)

*NOTE: (Of course it all depends on the inputed energy but lets say the input energy for the cubes is only 100J).

 

[(*steve*)]

Remember that 100J is an amount of energy.

If it takes 100J to lift 1 cube 1m, then you can't use the same 100J to lift 2 cubes 1m...

unless...

1) the second cube has no mass.
2) the sum of the masses of the 2 cubes is equal to the first.
3) you do this in a location where the force due to gravity is less
4) more energy comes from somewhere.

 

[timothy48342]

First of all, joules is energy, but it would be better to desrcibe a motor in power, which is energy per time unit. So, Joules per second for instance.

An imaginary motor might convert 110 Joules of electrical energy into 100 Joules of mechanicle energy per second with the other 10 Joules lost as heat or some other unrecoverable form. (110J/s in and 100J/s out and 10J/s lost)

(In general I think it is ok to simply say that some energy is "lost" as long as everyone understands that that energy is not destroyed or deleted.)

Now you want to consider what energy is "received by" the cube. I'm not sure how to reword that better. "Recieved" makes it sounds like the cube is an energy conatinor. I'm not saying that's wrong. I would say the cube gains some energy. It is important to note that the cube can some energy in more than one form.

Potential energy is the energy of a mass being suspended above the gravitational field of the earth. A rock sitting on a hill has more potential energy then tha same rock sitting at the base of the hill. Potential energy (U) is increased when it's height is increased. You can't state the absolute potential energy of a mass, but you can state the change in potential and calculate it. Change in U = mgh. (m is mass, g is a constant, h is change in height)

Kinetic energy (K) is based on velocity and mass. K = m * v^2 (mass times velocity squared.) Kenetic energy is the energy of motion.

Now I'm asuming in your example that you want all of the energy from the motor to be used up in lift the block vertically. That would mean that when it reaches a 1 meter height is has come to a stop. You not spending energy to stop is from moving, because that would be something else entirely. You spending energy to accelerate it upward and your spending only just enough so the it comes to a stop at 1 meter and all of the energy is in the form of potential. (no final motion, so nothing in the form of kinetic)

Thta's a very heavy block. But lets just say you've done that. The motor is doing all it can possibly do to lift this block. All of the 100J has been spent doing just that. The motor had nothing left over to give. Now the motor lifts 2 blocks over 1 meter. Wait! It can't.

If you want to say that the motor can lift both blocks, then I have to say that the motor had some left over energy(or power) that wasn't being completely utilized when lifting just 1 block.

If you want to switch over to using power measurements instead of energy. A motor might output 100J/s power and if It can lift 1 block over 1 meter in a specific amount of time using all of it's available power, then when lifting 2 blocks over 1 meter, it would take 2 times as much energy, so it would take the motor 2 times longer to do the job.

Now one more way to look at this. (I want to try and look at it as close as possible to what you described.) Let say that 100J of energy "goes into" or '"is received by" 1 block. Well, If is a very heavy block, then maybe it can't even be lifted 1 meter. (The potential energy to lift it 1 meter is it's mass earths ground-level gravitational constant g times 1 meter. One unit used for this is Newton-meters, but that is energy just like joules is energy.) Maybe moving it 1 meter is more than 100J. If it is a lightweight block, then the potential energy of lifting it 1 meter is less than 100J. Where does the extra energy go? We want to say that the block "has" that enregy, but it isn't potential energy. I must be kinetic energy, meaning that when the block reaches 1 meter, it is still moving. If it is a very lightweight block, then the potential energy is very small and so the kenetic energy must be quite large.

Imagine a powerful motor lifting a tiny weight as hard and as fast as it can. The weight goes flying way up in the air.

Now when you switch to using 2 blocks. Same thing. If they are too big, then motor can't do it. If they are small, then some energy is left over as kinetic energy. But since their are 2 blocks instead of 1, the potential energy is double. (mass is in the potential energy formula and your doubling the mass when you double the number of blocks) So there would be less energy left over of kinetic. That kinetic energy would be divided between the 2 blocks. So they would not be moving quite as fast as just the 1 block.

Of course there is a third situation that the motor can lift 1 block with some energy leftover and it can't left 2 blocks at all, but I don't think that 3rd situation help learn anything.

Bottom lin... You can't just ignore the mass of the blocks. I understand that you want it to be a general thing, but the mass plays an important roll in energy calculations.

--tim

 

[Moha99]

I understood why mass is an important unit in the equation but all of what I said is theoretical and couldn't really think of a number...

Now today a train passed infront of me and I was wondering... Now it has 1 engine that can supply enough mechanical power to move 100 carts... Looking at it, it had a lot of cargo! I sat there saying wow... One cart with fuel can get enough power to move all those carts.

Now what I wonder is mainly on rotational mechanical energy where for example that energy cart's 8 wheels or 6 wheels generate for those other 100 carts wheels.

But how is that...? If there was a train going on a still round with no up and downs and it generates maybe 1000J's of energy does every wheel on every cart besides the engine on get that 1000J transfered?(Ignoring the loss of energy just to understand the picture here) And with high momentum(of rotation) mass is decreased right?

Now this example is mainly on rotational energy because its kinda... Interesting

If you would like to apply an example with mass feel free it gives me more colors to pait this image perfectly.

 

[timothy48342]

Let's say that during some unknown amount of time a train has generated an amount of energy and successfully transfered 1000J of that to the train cars or carts behind it by pulling and accellerating them over level ground.

Where has that energy gone?

There is no potential energy change since there were no hills.
Ignoring any loses, the remainder must all be kinetic energy, but there are 2 types of kinetic energy, here.

One is the kind of kinetic energy we mentioned before. (I guess that would be the "normal" kind, but it's not called that. It's just called kinetic energy.)
The other type is Rotational Energy. It is actually a type of kinetic energy and it is sometimes called Rotational Kinetic Energy. It is still an energy of motion, but instead of everything moving all in one direction, everything moves in a circle.
(I don't want to say much about the formula, because it has been a while. I'm not sure.)

I'm not sure what you are asking about all the wheels. If 900J ended up being kinetic and 100J ended up being Rotational Kinetic, and there were 5 cars and 4 wheels per car for a total of 20 wheels and if each wheel was identicle, then each wheel would have 5J of rotational kinetic. 5J * 4 wheels * 5 cars = 100J

If that were the case and you lifted a car up in the air, the wheels would still be spinning and it would take 5J of energy to stop each wheel. Those numbers are just samples, we don't know the mass of the cars, or the size of the wheels or anything.

You say, "And with high momentum(of rotation) mass is decreased right?" Don't know what you mean. There is something called "moment of inertia" which basically describes how heavy something feels when you try to spin it., but that would have the opposite relation. High moment of inertia generally is associated with more mass, not less.

Not lets change the original question slightly:
Now let's say that an engine is generating energy and successfully transfering that to the train cars at a rate of 1000J per minute. Let's also say that the whole train has reached it's top speed and on level track.

Where is that energy going?
No chage in potential.
There is kinetic energy and rotational energy, but the are not chaning. They are not increasing since we said the train is already at top speed.
Ignoring loses, where is the rest of it going??

Here we have a problem. We can't ignore loses amymore. ALL OF THE ENERGY that the engine is transfering to the cars is being lost. Friction, heat, noise. If the engine runs out of fuel, it will immediately start to slow down.

In reality we couldn't ignore it before either. It is ok not to talk about the loses for the purpose of understanding the other types of energy, but in real life it happens.

Imagine a train is moving at a slow speed like 1 mph (or 1 kph, your choice.) and imagine it has run out of fuel and your job is to run behind it and and push to keep it moving and get to the next stop. You don't have to increase it's kinetic energy. You don't have to push it up a hill. Just keep it going the same speed. There is no way you can do it. It's not because it's heavy, it is becasue all those wheels turning and all those tiny bumps on the track add up and the frictional forces are way larger than you can produce by hand.)

I'm just saying it's ok to ingore friction as long as you remember that real life doesn't.

When it comes to mass, we don't alwasys have to give it a number, but we have to know that it is there. If we want to be very clear describing a problem about a block to be lifted, but we don't want to make up a number for the mass, we can say, "A motor lifts a block of unknown mass" or "a block of some mass." When the second block comes along we should say, "a second block that is the same mass as the first," or "a second block like the first."

It is even ok to not make any refernce to mass at all, but in your own mind, you have got to be thinking, "these blocks do have mass, I'm just not saying it out loud." That way when you add the second block into the problem, you don't fool yourself into thinking that mass doesn't play a role.


Have a great day!!
--tim

 

[Moha99]

First of all, joules is energy, but it would be better to desrcibe a motor in power, which is energy per time unit. So, Joules per second for instance.

An imaginary motor might convert 110 Joules of electrical energy into 100 Joules of mechanicle energy per second with the other 10 Joules lost as heat or some other unrecoverable form. (110J/s in and 100J/s out and 10J/s lost)

(In general I think it is ok to simply say that some energy is "lost" as long as everyone understands that that energy is not destroyed or deleted.)

Now you want to consider what energy is "received by" the cube. I'm not sure how to reword that better. "Recieved" makes it sounds like the cube is an energy conatinor. I'm not saying that's wrong. I would say the cube gains some energy. It is important to note that the cube can some energy in more than one form.

Potential energy is the energy of a mass being suspended above the gravitational field of the earth. A rock sitting on a hill has more potential energy then tha same rock sitting at the base of the hill. Potential energy (U) is increased when it's height is increased. You can't state the absolute potential energy of a mass, but you can state the change in potential and calculate it. Change in U = mgh. (m is mass, g is a constant, h is change in height)

Kinetic energy (K) is based on velocity and mass. K = m * v^2 (mass times velocity squared.) Kenetic energy is the energy of motion.

Now I'm asuming in your example that you want all of the energy from the motor to be used up in lift the block vertically. That would mean that when it reaches a 1 meter height is has come to a stop. You not spending energy to stop is from moving, because that would be something else entirely. You spending energy to accelerate it upward and your spending only just enough so the it comes to a stop at 1 meter and all of the energy is in the form of potential. (no final motion, so nothing in the form of kinetic)

Thta's a very heavy block. But lets just say you've done that. The motor is doing all it can possibly do to lift this block. All of the 100J has been spent doing just that. The motor had nothing left over to give. Now the motor lifts 2 blocks over 1 meter. Wait! It can't.

If you want to say that the motor can lift both blocks, then I have to say that the motor had some left over energy(or power) that wasn't being completely utilized when lifting just 1 block.

If you want to switch over to using power measurements instead of energy. A motor might output 100J/s power and if It can lift 1 block over 1 meter in a specific amount of time using all of it's available power, then when lifting 2 blocks over 1 meter, it would take 2 times as much energy, so it would take the motor 2 times longer to do the job.

Now one more way to look at this. (I want to try and look at it as close as possible to what you described.) Let say that 100J of energy "goes into" or '"is received by" 1 block. Well, If is a very heavy block, then maybe it can't even be lifted 1 meter. (The potential energy to lift it 1 meter is it's mass earths ground-level gravitational constant g times 1 meter. One unit used for this is Newton-meters, but that is energy just like joules is energy.) Maybe moving it 1 meter is more than 100J. If it is a lightweight block, then the potential energy of lifting it 1 meter is less than 100J. Where does the extra energy go? We want to say that the block "has" that enregy, but it isn't potential energy. I must be kinetic energy, meaning that when the block reaches 1 meter, it is still moving. If it is a very lightweight block, then the potential energy is very small and so the kenetic energy must be quite large.

Imagine a powerful motor lifting a tiny weight as hard and as fast as it can. The weight goes flying way up in the air.

Now when you switch to using 2 blocks. Same thing. If they are too big, then motor can't do it. If they are small, then some energy is left over as kinetic energy. But since their are 2 blocks instead of 1, the potential energy is double. (mass is in the potential energy formula and your doubling the mass when you double the number of blocks) So there would be less energy left over of kinetic. That kinetic energy would be divided between the 2 blocks. So they would not be moving quite as fast as just the 1 block.

Of course there is a third situation that the motor can lift 1 block with some energy leftover and it can't left 2 blocks at all, but I don't think that 3rd situation help learn anything.

Bottom lin... You can't just ignore the mass of the blocks. I understand that you want it to be a general thing, but the mass plays an important roll in energy calculations.

--tim

Let's say that during some unknown amount of time a train has generated an amount of energy and successfully transfered 1000J of that to the train cars or carts behind it by pulling and accellerating them over level ground.

Where has that energy gone?

There is no potential energy change since there were no hills.
Ignoring any loses, the remainder must all be kinetic energy, but there are 2 types of kinetic energy, here.

One is the kind of kinetic energy we mentioned before. (I guess that would be the "normal" kind, but it's not called that. It's just called kinetic energy.)
The other type is Rotational Energy. It is actually a type of kinetic energy and it is sometimes called Rotational Kinetic Energy. It is still an energy of motion, but instead of everything moving all in one direction, everything moves in a circle.
(I don't want to say much about the formula, because it has been a while. I'm not sure.)

I'm not sure what you are asking about all the wheels. If 900J ended up being kinetic and 100J ended up being Rotational Kinetic, and there were 5 cars and 4 wheels per car for a total of 20 wheels and if each wheel was identicle, then each wheel would have 5J of rotational kinetic. 5J * 4 wheels * 5 cars = 100J

If that were the case and you lifted a car up in the air, the wheels would still be spinning and it would take 5J of energy to stop each wheel. Those numbers are just samples, we don't know the mass of the cars, or the size of the wheels or anything.

You say, "And with high momentum(of rotation) mass is decreased right?" Don't know what you mean. There is something called "moment of inertia" which basically describes how heavy something feels when you try to spin it., but that would have the opposite relation. High moment of inertia generally is associated with more mass, not less.

Not lets change the original question slightly:
Now let's say that an engine is generating energy and successfully transfering that to the train cars at a rate of 1000J per minute. Let's also say that the whole train has reached it's top speed and on level track.

Where is that energy going?
No chage in potential.
There is kinetic energy and rotational energy, but the are not chaning. They are not increasing since we said the train is already at top speed.
Ignoring loses, where is the rest of it going??

Here we have a problem. We can't ignore loses amymore. ALL OF THE ENERGY that the engine is transfering to the cars is being lost. Friction, heat, noise. If the engine runs out of fuel, it will immediately start to slow down.

In reality we couldn't ignore it before either. It is ok not to talk about the loses for the purpose of understanding the other types of energy, but in real life it happens.

Imagine a train is moving at a slow speed like 1 mph (or 1 kph, your choice.) and imagine it has run out of fuel and your job is to run behind it and and push to keep it moving and get to the next stop. You don't have to increase it's kinetic energy. You don't have to push it up a hill. Just keep it going the same speed. There is no way you can do it. It's not because it's heavy, it is becasue all those wheels turning and all those tiny bumps on the track add up and the frictional forces are way larger than you can produce by hand.)

I'm just saying it's ok to ingore friction as long as you remember that real life doesn't.

When it comes to mass, we don't alwasys have to give it a number, but we have to know that it is there. If we want to be very clear describing a problem about a block to be lifted, but we don't want to make up a number for the mass, we can say, "A motor lifts a block of unknown mass" or "a block of some mass." When the second block comes along we should say, "a second block that is the same mass as the first," or "a second block like the first."

It is even ok to not make any refernce to mass at all, but in your own mind, you have got to be thinking, "these blocks do have mass, I'm just not saying it out loud." That way when you add the second block into the problem, you don't fool yourself into thinking that mass doesn't play a role.


Have a great day!!
--tim

hahaha thanks Tim really helped me out there.

But in you're previous reply you also included the 1 meter/second things and even in the second one there is a 1 something per/ something. Point of it all is a motor for example is rated fro it rpm's(revolutions per minute) since its making for example 1000RPMS/Min we can add the idea the the motor is using 100J to deliver 1000rpm/min. But I isn't the motor constantly taking 100J from the power source /per min? That energy is given out and lost in the process.

and + a motor can lift double its weight so that 1000RPM's that was supplied from a source can rotate something double its weight.

 

[timothy48342]

Using a unit per time instead of a unit by itself:
If you said you had a motor that could go 100 revoutions, I would wonder if that is 100 revolutions per second or per minute. I would then ask, "after this motor makes goes 100 revoutions, can it go 100 revoutions again or do we just throw it away?"

Saying "100 revoutions" by itself doesn't mean anything. Although you could say "100 revoutions during an experiment" or "100 revoutions over an unspecified time period" and that's fine, but to say it by itself leaves the reader asking 100 revoutions... per what?

The same with energy. If a motor can produce 1000J, what does that mean? 1000J per second, minute, year?

I think you meant 1000J during an experiment, which is fine, but I couldnt' tell for sure, so I wanted to clarify.

Energy drawn from a power sourcce versus energy delivered usefully:
IF a motor is drawing a constant amount of energy per minute and spinning freely, then all of the energy is wasted. If it connected to a light load, then its likely that a lot will be lost. It is also possible that the energy (or power) drawn will chage with the load. When spinning freely the motor might take a lot less power compared to when lifting a heavy load. (There might be certain types of motors that have a constant power draw regardless of load. I'm not sure.)

Can a motor lift double it's weight?
Many can. Many can lift way more than that? A poorly designed motor can't lift much at all. I know one guy that designed a motor that didn't even work! So to use double it's weight as a rule of thumb is not right. Who told you that? I would say that most motors would be able to lift much more than double their weight.

--tim

 

[Moha99]

Using a unit per time instead of a unit by itself:
If you said you had a motor that could go 100 revoutions, I would wonder if that is 100 revolutions per second or per minute. I would then ask, "after this motor makes goes 100 revoutions, can it go 100 revoutions again or do we just throw it away?"

Saying "100 revoutions" by itself doesn't mean anything. Although you could say "100 revoutions during an experiment" or "100 revoutions over an unspecified time period" and that's fine, but to say it by itself leaves the reader asking 100 revoutions... per what?

The same with energy. If a motor can produce 1000J, what does that mean? 1000J per second, minute, year?

I think you meant 1000J during an experiment, which is fine, but I couldnt' tell for sure, so I wanted to clarify.

Energy drawn from a power sourcce versus energy delivered usefully:
IF a motor is drawing a constant amount of energy per minute and spinning freely, then all of the energy is wasted. If it connected to a light load, then its likely that a lot will be lost. It is also possible that the energy (or power) drawn will chage with the load. When spinning freely the motor might take a lot less power compared to when lifting a heavy load. (There might be certain types of motors that have a constant power draw regardless of load. I'm not sure.)

Can a motor lift double it's weight?
Many can. Many can lift way more than that? A poorly designed motor can't lift much at all. I know one guy that designed a motor that didn't even work! So to use double it's weight as a rule of thumb is not right. Who told you that? I would say that most motors would be able to lift much more than double their weight.

--tim

I meant by "1000rpm/Revolutions per minute" and the energy source is constant and keeps feeding that monster always!

So if a motor can handle double its mass and possibly more thats just amazing.

One thing to ask though the shaft lets say we have a 1killo motor what will that motors shaft be weighed at? would it be very light? or very heavy? maybe it would be 0.200killo or something.

 

[davenn]

hahaha thanks Tim really helped me out there.

But in you're previous reply you also included the 1 meter/second things and even in the second one there is a 1 something per/ something. Point of it all is a motor for example is rated fro it rpm's(revolutions per minute) since its making for example 1000RPMS/Min we can add the idea the the motor is using 100J to deliver 1000rpm/min. But I isn't the motor constantly taking 100J from the power source /per min? That energy is given out and lost in the process.

the power of the motor isnt rated by its RPM. Its power is in its horsepower rating and thats irrelevent of its its a pertol, diesel or electric motor. its the torque that the motor that indicates its turning power.

and + a motor can lift double its weight so that 1000RPM's that was supplied from a source can rotate something double its weight.

a motor that can turm at 1000 RPM isnt necessiarly stronger than a motor that turns at 500RPM
You often find that very high torque (power) motors have a much slower RPM...
just look at the difference between a Diesel and a petrol engine. The diesel generates much more power at lower revs, hence why they are used in heavy machinery

Dave

 

[BobK]

Why is it surprising to you that a motor can lift more than its weight? Have you never driven a car up a hill?

Bob