Yesterday I was moving 40 lb. bags of humus/manure around the yard, four bags at a time. It took a substantial effort to start the wheel barrow moving, but less so to keep it moving.
Objects in motion tend to stay in motion. It’s a phenomenon that I’m sure everyone has noticed many times . If you try to push a car, the hardest part is getting it going from a standstill.
Here’s the question:
Does it take more energy to accelerate the car (or wheelbarrow) from zero to 1 mph than it does to accelerate it from 1 mph to 2 mph? Does it take a lot more? If so, why?
I have some guesses, but there’s no need to air them.
In every day life, no. To accelerate something of fixed mass, will take the same amount of energy, to accelerate it per unit of time/velocity. Practically speaking.
Bring relativity into it, and the closer you get to the speed of light, the more mass it will gain, and the more energy it’ll take to accelerate (this is why traveling up to / faster than the speed of light is considered impossible, because that is the point where it would gain infinite mass and take infinite energy to keep accelerating). But this isn’t really noticeable until you get into incredibly fantastic speeds.
I’ll lay low and let a real physicist explain more, or correct any misconceptions I may have.
On the earth, you’ll be fighting against air resistance, depending on the situation. Obviously, the faster you go, the higher the resistance will get, and the more energy it’ll take to keep accelerating (this is what astronauts use as a brake for reentry, hence the heat shields).
Well, what cmyk is describing is really the ideal physics test question world where we assume frictionless surfaces and such. And in such cases the energy used from 0 mph to 1 mph is the same as from 1 mph to 2 mph or 6,000 mph to 6,001 mph.
In the real world, there is friction, and typically those friction forces are greater when the object is at rest than when the object is moving (static friction versus kinetic friction). So going from 0 to 1 will result in expending energy to overcome additional friction forces that you won’t have going from 1 to 2. How much more energy is required depends greatly on a vast number of variables.
No, that’s not true. I suspect you-all (as well as Baal Houtham) are conflating energy with force. From the perspective of energy, it takes far more energy to accelerate an object from 6000mph to 6001mph than it does to accelerate it from 1mph to 2mph. That’s because the kinetic energy of an object is 1/2mv[sup]2[/sup], and 6001[sup]2[/sup] - 6000[sup]2[/sup] is far more than 2[sup]2[/sup] - 1[sup]2[/sup].
And yes, this is a large part of what’s going on. The amount of force (note the term here) required to juuuuuust get the wheelbarrow moving is greater than the amount of force it takes to push it at a constant 2mph, because the static friction is higher than the dynamic friction. Just why that is true gets rather complex.
It’s also possible that Baal’s wheelbarrow tends to park itself in a slight dip in the grass, requiring noticeably more force to push it out; force that wouldn’t be required once the wheelbarrow’s moving and jouncing in and out of the slight dips. I’m not sure how significant that effect is, if it is at all.
It is not the same amount of energy to go from 0 to 1 MPH as from 1 to 2, even in the textbook idealized Newtonian world. Energy is proportional to the square of the speed, so it would actually take three times as much energy to go from 1 to 2 as it does from 0 to 1.
Ignoring relativistic considerations, at 2 MPH, your wheelbarrow (or any other object) has four times the kinetic energy it possesses at 1 MPH, quite the opposite of what your question presupposes. The equation for kinetic energy is:
E = 0.5 * m * V^2
Where:
E = kinetic energy
m = object mass
V = object velocity
So if you 2x your speed, you 4x your kinetic energy.
That’s the simplest physics approach. In real life, there may be “stiction” on moving surfaces that requires a bit of extra oomph to break loose initially. Bearings, seals, brake pads on rotors, and so on. If you were dropping items into your wheelbarrow, you may have caused the wheel to embed itself in a divot in the ground, which will require a bit of extra oomph to push out of initially.
Rolling resistance and wind/bearing drag will tell you how fast your object will decelerate once up to speed, or how much power you will require to maintain a given speed. but if your object is accelerated with reasonable alacrity, then the basic kinetic energy equation will tell you what’s going on.
Also keep in mind that air resistance incresses as the square of the velocity. In other words the faster you go the more energy is required to accelerate.
I’m imagining a 1 meter cube of aluminum floating in LEO. If I wanted to accelerate that cube from 1 to 2 mph, using my trusty rocket-pack, wouldn’t it take the same amount of fuel that it took to accelerate it from 0 - 1 mph. (ignoring relativistic effects)?
Yes, it would take the same amount of fuel. This is because a rocket engine is inefficient at low vehicle speeds (where “low” is compared to the exit velocity of the rocket’s exhaust).
Suppose the exhaust gases leave the rocket’s engine bell at 2000 MPH relative to the bell; this will be true regardless of vehicle speed.
When you are accelerating from 0 to 1 MPH, the exhaust (on average) ends up with a final absolute velocity of 1999.5 MPH, and a particular amount of kinetic energy.
When you accelerate from 1 to 2 MPH, the rocket exhaust expelled during this event has a final absolute velocity of 1998.5 MPH, and less kinetic energy than the first batch of exhaust did; this missing kinetic energy has been given to the vehicle instead, providing the extra energy needed to get from 1 to 2 MPH that you didn’t need when accelerating from 0 to 1 MPH.
Yes, that was my point. Energy and force are different things. They’re related, yes, in that force is (or in general can be) a component of energy, but they’re different quantities.
Some of the confusion arises from the casual usage of both “force” and “energy” to mean “effort,” as in, “it takes a lot of effort/energy/force to push this wheelbarrow across the lawn.” However, the technical definitions of "force " and “energy” are precise and differentiated, and when one frames a technical question like Baal Houtham’s, the answer depends on the definitions.
zut: As I read the OP, I figured what he meant was would it take the same amount of work/effort to push the wheelbarrow to 1-2 mph, that it took to push it from 0-1 mph. Alas, I see I’m in over my head!
35 years ago my job in this Iron Ore Mine.
At one time I was a “Car Rider”. the job was to spott empty 70 ton ore cars under the loading pocket and then ride them down the track coupling them together to make a train.
To start the cars moving, loaded or empty, it was the most common practice to push from between 2 cars with back against one and one leg pushing the other. This worked 90% of the time. (there was a slight down slope)
When that didn’t work, we used a teaser (pinch bar) to get the car moving.
Just like with a wheelbarrow, you just put your back into it.
I think this question is much trickier than just referring to behavior in the limit. I think you have to account for the kinetic energy in the rockets and their remaining fuel and, significantly, their exhausts (which are clouds with distributed kinetic energies).
Otherwise you can have this funny situation where you buy two identical rockets with ten hour motors, and attach each to a winch, and start both rockets with the winch brakes off. After a minute you adjust one of the winch brakes so that that its force exactly counters the rocket thrust. After a second minute you adjust the other winch brake so that that one counters its rocket thrust. The second rocket accelerated for twice as long and is now going twice as fast as the first. Now you have very nearly ten hours to wait, during which the second rocket appears to be doing twice as much work as the first one.
The old WWI four stacker destroyers had four fire rooms. With two of them in operation they could make 26 knots. Bringing the other two on line would get them up to 32 knots. So basically 100% more power gave 23% more speed.
Wait, I was always taught that there was no absolute frame of reference from which to measure speed (obviously, we’re talking vacuum, yada, yada, here). So, bearing that in mind, I would have thought that in that vacuum, the speed of the rocket + cube wouldn’t matter - going from X to X+1 mph is the same thing. And, yeah, forget about rocket fuel and weight and all that.
ETA, in the wheelbarrow case, you also have to move yourself faster pushing it at 2 mph, which requires some energy.
It doesn’t just appear to be doing twice as much work, it actually is doing twice as much work. As stated in my earlier post, a rocket engine is more efficient at higher speeds.
Thrust x speed = power delivered to vehicle. Rocket provides constant thrust; if it’s moving at higher speed, it’s delivering more power to the vehicle.
In your scenario, that second winch brake will heat up twice as quickly as the first winch brake.
