B Why higher speeds need more power if backward force is the same?

[gen x]

Power = Force v Speed

Power of my horse = 104kgx9.81m/s^2 x 0.732m/s = 1HP =746W

Force/tension in rope stay the same if horse run at 0.73m/s or at 15m/s, so why then horse need to be more powerfull to pull at higher speed even if backward force at him(rope tension) stay the same?
I understand that if I increase weight, it is hrader for horse to pull at higher speed because now is backward force increased, but don't understand why is harder to pull at higher speed if weight(backward force) is the same.

 

[Lnewqban]

One thing is work performed by the horse, and a different one is the time in which that work is done.
Less time requires greater power.
Olympic games are mainly about that fact.
Your feet excert the same force on the pedals of your bicycle to climb the same hill, either slowly or quickly, but it is only so fast that you can pedal.

 

[gen x]

One thing is work performed by the horse, and a different one is the time in which that work is done.
Less time requires greater power.

Imagine if car travel in different medium so at 100km/ and 200km/h is same drag force at car.
It turns out that car that travel at 200km/h need more power only because it travel at faster speed, even medium drag is the same, isn't that wierd?
That show me, my physics/life intuition is totally wrong.

 

[PeroK]

Imagine if car travel in different medium so at 100km/ and 200km/h is same drag force at car.
It turns out that car that travel at 200km/h need more power only because it travel at faster speed, even medium drag is the same, isn't that wierd?
That show me, my physics/life intuition is totally wrong.

Consider an object falling in vacuum under a constant gravitational force. The rate that kinetic energy is gained by the object increases with time, as the object accelerates.

Constant force, therefore, implies increasing power.

 

[gen x]

Olympic games are mainly about that fact.
Your feet excert the same force on the pedals of your bicycle to climb the same hill, either slowly or quickly, but it is only so fast that you can pedal.

When I decrease F and use higer gears ratio, I can incease my speeed, but my power still remain the same.

Consider an object falling in vacuum under a constant gravitational force. The rate that kinetic energy is gained by the object increases with time, as the object accelerates.

Constant force, therefore, implies increasing power.

Is my example with horse counter intuitive to you as well or not?

 

[PeroK]

Is my example with horse counter intuitive to you as well or not?

There are a lot of people who think that applied force remains constant as speed increases. They see applied force as the fundamental quantity that can be maintained as speed increases. E.g. in pushing an object, or accelerating a vehicle such as a bike or car. They believe that acceleration only becomes more difficult because of increasing resistance forces, which eventually equal the constant accelerating force.

In fact, leaving aside certain complications, it's the available power that is the constant. As speed increases, the same power yields a smaller accelerating force. Even without any resistance forces, acceleration will decrease. And, eventually, the accelerating force reduces to the point where even a small resistance force will equal it.

Also, as speed increases, resistance forces tend to increase, which makes the reduction in acceleration with speed even more dramatic.

I'm not sure whether what I know about forces, energy and power in this context is intuitive (in whatever sense that means). At some point, I learned the physics of acceleration, and that's all I can say.

 

[gen x]

There are a lot of people who think that applied force remains constant as speed increases. They see applied force as the fundamental quantity that can be maintained as speed increases. E.g. in pushing an object, or accelerating a vehicle such as a bike or car. They believe that acceleration only becomes more difficult because of increasing resistance forces, which eventually equal the constant accelerating force.

In case of bike, lets say my foot push on pedals only in half circle, when I am above pedal so my weight ( mg) make force on pedals. How my force on pedals decrease with speed if mg is constant?

 

[Dale]

Walk up some flights of stairs. Then the next time run up the same stairs. Which makes you more sweaty?

 

[gen x]

Walk up some flights of stairs. Then the next time run up the same stairs. Which makes you more sweaty?

When I run is harder, but when I run I use more power , my leg rpm is higher and my force push on each stair is higher.

 

[PeroK]

In case of bike, lets say my foot push on pedals only in half circle, when I am above pedal so my weight ( mg) make force on pedals. How my force on pedals decrease with speed if mg is constant?

A bike is a complicated machine. As the pedals speed up, it becomes harder to maintain the force. Without changing gear, you end up feeling very little resistance from the pedals. It doesn't matter whether you try to stand up on the pedals - that doesn't work if you are already going fast.

Especially if you are on a downhill section, you can hardly generate any force on the pedals at all.

The gears are important in terms of allowing you to generate more force on the road from the force on the pedal. That allows you to keep accelerating when it would be impossible on a simple bicycle without higher gears.

 

[DaveC426913]

In fact, leaving aside certain complications, it's the available power that is the constant. As speed increases, the same power yields a smaller accelerating force. Even without any resistance forces, acceleration will decrease. And, eventually, the accelerating force reduces to the point where even a small resistance force will equal it.

This seems wrong. I am thinking about a rocket in space. Ignoring the change in mass due to fuel loss, surely a rocket accelerates at a constant rate for a constant appliciation of force.

What am I missing?

(Have I forgotten to factor in the max velocity being limited to twice exhaust velocity?)

 

[kuruman]

Force/tension in rope stay the same if horse run at 0.73m/s or at 15m/s, so why then horse need to be more powerfull to pull at higher speed even if backward force at him(rope tension) stay the same?

Here is an easy to understand example. There are 20 boxes of books, each of mass 20 kg, that need to be moved to an apartment 10 meters above ground. There is no elevator. A powerful weightlifter might stack 5 boxes on top of each other and finish the job in 4 trips and 20 minutes total. An average college student will probably take one box at a time and finish the same job in 2 hours (with breaks). The work done against gravity is the same in both cases. However, the work done per unit time (power) is higher in the first case because the weightlifter, being more powerful than the college student, can do the same work in less time.

In your example you have a horse lifting a weight against gravity but the idea is the same. If the horse moves at a higher speed, it lifts the weight by a fixed amount, say 2 m, in less time and therefore has to use more power.

 

[PeroK]

This seems wrong.

It's not wrong, but a rocket is a special case. See below.

I am thinking about a rocket in space. Ignoring the change in mass due to fuel loss, surely a rocket accelerates at a constant rate for a constant appliciation of force.

Everything accelerates at a constant rate for a constant application of force.

What am I missing?

There are two ways to look at a rocket:

1) Assuming it is a closed system, then it cannot accelerate. What it can do is split apart and redefine a reducing fraction of the original system as the "rocket". There is no external force, as such. And the centre of mass of the original body of mass does not accelerate.

2) If we allow the rocket to be continually redefined, then it is capable of acceleration with the expellent providing an external force that is constant (even increasing) and there is not a constant power. It's continually redfining what is meant by "the rocket" that makes it a special case.

 

[DaveC426913]

It's not wrong, but a rocket is a special case. See below.

I chose a rocket because it is the simplest case. Surely, every other example is the special case.

Everything accelerates at a constant rate for a constant application of force.

OK, that is what I expected.

But it seems to fly directly in the face of what I quoted you saying.

2) If we allow the rocket to be continually redefined, then it is capable of acceleration with the expellent providing an external force that is constant (even increasing) and there is not a constant power. It's continually redfining what is meant by "the rocket" that makes it a special case.

You are talking about the rocket's decreasing mass. I specifically ignored that.

(Say my rocket is so big that its fuel loss for the test is miniscule by comparison, and so efficient that applies virtually all its energy as exhaust.)

Are you saying that such an ideal rocket - even in principle - will not maintain a constant acceleration for constant force?

 

[kuruman]

This seems wrong. I am thinking about a rocket in space. Ignoring the change in mass due to fuel loss, surely a rocket accelerates at a constant rate for a constant appliciation of force.

What am I missing?

(Have I forgotten to factor in the max velocity being limited to twice exhaust velocity?)

The rocket equation in the absence of external forces is $$m\frac{dv}{dt}=(u-v)\frac{dm}{dt}$$where
$m=~$ instantaneous mass of the rocket + unspent fuel
$v=~$ instantaneous velocity of the rocket relative to an inertial frame
$u=~$ instantaneous velocity of the exhaust gas relative to the same inertial frame

You can see that when $v=u$ the acceleration is zero.

 

[Baluncore]

The rate of energy flow, that an engine can convert, is its power rating.
Kinetic Energy; E_k= ½⋅m⋅v²
Velocity; v = √ ( 2 ⋅ E_k / m )
As energy flows at a fixed rate, the velocity can only rise at a reducing rate.

 

[DaveC426913]

The rocket equation in the absence of external forces is $$m\frac{dv}{dt}=(u-v)\frac{dm}{dt}$$where
$m=~$ instantaneous mass of the rocket + unspent fuel
$v=~$ instantaneous velocity of the rocket relative to an inertial frame
$u=~$ instantaneous velocity of the exhaust gas relative to the same inertial frame

You can see that when $u=v$ the acceleration is zero.

OK, so you are confirming that the exhaust velocity is the limiting factor for rockets.

 

[gmax137]

Power of my horse = 104kgx9.81m/s^2 x 0.732m/s = 1HP =746W

Force/tension in rope stay the same if horse run at 0.73m/s or at 15m/s, so why then horse need to be more powerfull to pull at higher speed even if backward force at him(rope tension) stay the same?

EDIT: @gen x , forget about rockets for now. Just think about your horse \EDIT

Let's say the horse pulls for 10 meters. What is the work done?

1020 N * 10 m = 10,200 Nm

if the horse's velocity is 1 m/sec, what is the power?

power is work/time, and time is distance/velocity, so

$$power = \frac {10,200 Nm} {\frac {10 m}{1 m/s}} = 1020 \frac{Nm}{sec}$$

if the horse's velocity is 2 m/sec, what is the power?
$$power = \frac {10,200 Nm} {\frac {10 m}{2 m/s}} = 2040 \frac{Nm}{sec}$$

as @kuruman said,

In your example you have a horse lifting a weight against gravity but the idea is the same. If the horse moves at a higher speed, it lifts the weight by a fixed amount, say 2 m, in less time and therefore has to use more power.

 

[gen x]

This seems wrong.

Car acceleration decrease when speed rise (even if aero/tire drag is zero), because thrust force decrease.
Torque at wheels are smaller and smaller as you shift gears, gear ratios become lower each shift

 

[kuruman]

OK, so you are confirming that the exhaust velocity is the limiting factor for rockets.

Yes. In the inertial frame where the rocket is initially at rest and no fuel has been spent, the total momentum is zero and must remain zero at all subsequent times. In that frame, the very first $dm$ of spent fuel has speed $u_0$ and the accelerating rocket cannot move faster than that in the opposite direction.

 

[PeroK]

I chose a rocket because it is the simplest case. Surely, every other example is the special case.

That doesn't make a lot of sense. Running, bikes, cars, trains, aircraft, boats etc. can't all be the special case. Rocket propulsion is the special case in this context.

But it seems to fly directly in the face of what I quoted you saying.

No it doesn't. It's the difficulty in maintaining a constant force that is the problem. To maintain a constant force, you need an increasing power supply.

You are talking about the rocket's decreasing mass. I specifically ignored that.

You can't ignore it!

(Say my rocket is so big that its fuel loss for the test is miniscule by comparison, and so efficient that applies virtually all its energy as exhaust.)

The question is how to maintain a constant acceleration. If the rocket starts as N particles, then those N particles cannot accelerate purely by an internal engine. This is an important observation, that cannot be neglected.

The way to achieve "acceleration" is by ejecting a particle in one direction, redefining the remaining N-1 particles as the rocket, which has now accelerated.

That would also work on, say, a train on smooth tracks. You could throw mass out the back of the train and have makeshift rocket propulsion. This is the exception, because trains don't work like that. They work by the wheels pushing on the track, which provides the external force.

Are you saying that such an ideal rocket - even in principle - will not maintain a constant acceleration for constant force?

You've completely misunderstood what I've said.

 

[PeroK]

This seems wrong.

There are a lot of people who think that applied force remains constant as speed increases. They see applied force as the fundamental quantity that can be maintained as speed increases.

 

[wrobel]

I do not understand such questions. There is a definition: Power=$W=(\boldsymbol F,\boldsymbol v)$.
There is a theorem:
$$\dot T=W,\quad T=\frac{1}{2}m|\boldsymbol v|^2.$$
What else?

 

[jack action]

Say, your horse needs to eat 1 kg of hay (that's where it gets its energy) every 60 m it pulls the 1000 N weight.

If it goes at 1 m/s, every 1 minute the horse will have to eat its 1 kg of hay. After 1 hour, it will have covered 3600 m and have eaten 60 kg of hay. So 60 kg of hay per hour is required.

And if it goes 2 m/s, the horse will eat its 1 kg of hay every 30 seconds. After 1 hour, it will have covered 7200 m and have eaten 120 kg of hay. So 120 kg of hay per hour is required. For the same period of time, it requires more food (energy).

But if you have only 60 kg of hay available, if the horse goes at 2 m/s, it will still cover 3600 m, but it will take only 30 minutes.

Why the horse doesn't go at 100 m/s and do the same work faster? Because its digestive system couldn't handle converting 1 kg of hay to mechanical energy within 0.6 second. Every machine has its limit, and that is what power measures.

So, it is not harder to pull the 1000 N for a given distance at faster speeds. But you do need a machine capable of delivering the energy required faster. There is usually heat wasted to deal with during energy conversion, and cooling is then harder when it's done faster, thus why the horse will sweat more at higher power level.

Same difference with a car, where hay is replaced with fuel.