Need Basic Physics Help

[KrazyHorse]

In other words, you don't have to increase force to maintain the velocity at higher and higher levels, until the point where drag from the atmosphere comes into play.

[snoopy369]

Really? Hmmm, obviously confused then. I thought friction increased with velocity (not a lot, but a little).

In that case, then, if you add some force beyond that of friction, the cart will continue accelerating and never reach equilibrium, unless you factor in air resistance (drag).

[KrazyHorse]

In that case, then, if you add some force beyond that of friction, the cart will continue accelerating and reach equilibrium, unless you factor in air resistance (drag).

??? Please rewrite this so it makes sense.

[snoopy369]

Yeah, forgot my basic physics a bit there It increases for CARS, because the heat causes the tires to soften and more of the tire comes into contact with the road, but that's not relevant to the example. Oh well.

[Space05us]

In other words, you don't have to increase force to maintain the velocity at higher and higher levels, until the point where drag from the atmosphere comes into play.

Ignoring all forces other than the push and friction from the object's contact with the ground.

If it takes 1N at low speed to achieve dynamic equilibrium, then at very high speed it would also take 1N to achieve dynamic equilibrium?

[snoopy369]

In that case, then, if you add some force beyond that of friction, the cart will continue accelerating and never reach equilibrium, unless you factor in air resistance (drag).

??? Please rewrite this so it makes sense.

I left out a word that I thought but didn't type

In fake physics terminology: If you apply a constant force F to the cart that is greater than the force of friction Ff, the velocity V of the cart will increase at a constant rate, if you disregard air resistance.

[snoopy369]

Ignoring all forces other than the push and friction from the object's contact with the ground.

If it takes 1N at low speed to achieve dynamic equilibrium, then at very high speed it would also take 1N to achieve dynamic equilibrium?

If I read what he wrote correctly, then yes, that's what KH is saying (friction is effectively a constant force, so if 1N is the force of friction, you need to apply a constant 1N at any given speed to maintain a constant speed).

[KrazyHorse]

If it takes 1N at low speed to achieve dynamic equilibrium, then at very high speed it would also take 1N to achieve dynamic equilibrium?

Yes. The thing to note is that force and power are two vastly different things.

The power (energy per unit time) you (as the pushing force) are putting into the system is force times velocity. So even though the forces are the same, the power is very different.

[snoopy369]

OK, now I'm confused. I know that power and force are different, but what exactly do you mean there? Do you mean it's harder [takes more energy over time] to apply 1N of force at 100m/s than at 1m/s?

[KrazyHorse]

Yes, obviously. 100 times as much.

[snoopy369]

OK, good. Just wasn't sure if you were implying something more.

[KrazyHorse]

Your next question is probably going to be one about galilean relativity (this is a mildly insightful question)

Imagine you're on a train traveling at 99m/s to the right and you push a box with 1N of force at 1m/s (relative to the train) to the right. From the perspective of an observer on the train, this is the same as pushing a box at 1m/s with force 1N if the train wasn't moving at all; in other words, you have to provide 1W of power to push the box

From the perspective of an observer outside the train (standing on the ground), the situation is more complicated:

1) He sees you pushing the box with force 1N at 100m/s.
2) He sees the train pushing your feet with force 1N at 99m/s (each foot is at rest relative to the train as it pushes)

So this observer says to himself "the guy on the train is putting 100W into the box and getting back 99W from the train, so he's putting in 1W himself".

[Dry]

Really? Hmmm, obviously confused then. I thought friction increased with velocity (not a lot, but a little).

In that case, then, if you add some force beyond that of friction, the cart will continue accelerating and never reach equilibrium, unless you factor in air resistance (drag).

Kinetic friction is proportional to the contact surface, the weight of the object 'above' and the chemical bonding between the surfaces.
It has nothing to do with velocity.

For drag... well it's a little more complex, it depends on linear or turbulent flux, blah-blah-blah...

[KrazyHorse]

Kinetic friction is proportional to the contact surface

No.

[Dry]

From the perspective of an observer outside the train (standing on the ground), the situation is more complicated:

1) He sees you pushing the box with force 1N at 100m/s.
2) He sees the train pushing your feet with force 1N at 99m/s (each foot is at rest relative to the train as it pushes)

So this observer says to himself "the guy on the train is putting 100W into the box and getting back 99W from the train, so he's putting in 1W himself".

If the guy is a physicist, he just notice that the guy on the train is in an inertial frame of reference, so he can do his calculations from the observer on the train point of view.
And the same maybe applied to the space potato and the 2 gremlins. You can see it move at constant 1m/s in some inertial frame of reference, but you can also see it not moving at all in another, or see it move at 100m/s in a third one.
Makes no difference.