Why does it take more energy to accelerate faster objects?

[yellowcakepie]

Let's say we have a 1kg object and the reference point is the ground.

Accelerating it from 0 to 1 m/s takes 0.5*1*1^2 - 0 joules, which is 0.5 joules.

But accelerating it from 10 to 11 m/s takes 0.5*1*11^2 - 0.5*1*10^2 joules, which is 10.5 joules.

This doesn't make much sense to me as the increase in velocity was the same. Why can't an application of 0.5 joules increase the velocity by 1 m/s at any velocity?

 

[fresh_42]

Let $E(v)$ be the kinetic energy at velocity $v$. Now what is $E(v+\Delta v)-E(v)$ and what is $E(0+\Delta v)-E(0)\,$? Both increased by $\Delta v$. Do you see the difference?

 

[sophiecentaur]

The extra energy to accelerate (relative to the Earth frame of reference) reflects, exactly, the extra damage the object would do if it hit a 'brick wall. Kinetic Energy is not absolute. It is frame dependent.
The calculations you have done look ok to me. The only thing wrong is that you are letting your intuition get in the way of accepting those results.

 

[DrGreg]

Why can't an application of 0.5 joules increase the velocity by 1 m/s at any velocity?

Clue: work done by a force is force × distance, not force × time.

 

[fresh_42]

Just a note.

In everyday life, it is important to be aware of this difference in case of deceleration. Calculate the energy which is necessary to slow down a car from 160 km/h to 100 km/h in comparison to 60 km/h to a full stop. It is worth to keep this in mind.

 

[pixel]

The quick answer to OP's question is that kinetic energy is not linear in velocity but is proportional to the square of the velocity. So the real question is why this is so. See if the first answer here helps: https://www.quora.com/Why-does-kinetic-energy-increase-as-velocity-squared

 

[yellowcakepie]

Let $E(v)$ be the kinetic energy at velocity $v$. Now what is $E(v+\Delta v)-E(v)$ and what is $E(0+\Delta v)-E(0)\,$? Both increased by $\Delta v$. Do you see the difference?

Yes, but that was the phenomenon that I was describing in my original post.

The quick answer to OP's question is that kinetic energy is not linear in velocity but is proportional to the square of the velocity. So the real question is why this is so. See if the first answer here helps: https://www.quora.com/Why-does-kinetic-energy-increase-as-velocity-squared

I understand how mgh = 1/2mv^2 because of the decreased amount of time spent in a certain distance due to increased speed.

But what if we have a 100% efficient heat-to-kinetic energy converter accelerating our object by taking thermal energy from water? Work would be equal to -mcΔT. That means the water would have to be cooled down much more to accelerate from 10 to 11 m/s than 0 to 1 m/s? That seems wrong.

 

[jbriggs444]

But what if we have a 100% efficient heat-to-kinetic energy converter accelerating our object by taking thermal energy from water? Work would be equal to -mcΔT. That means the water would have to be cooled down much more to accelerate from 10 to 11 m/s than 0 to 1 m/s? That seems wrong.

By glossing over the details of how the acceleration is accomplished, you've removed the important stuff.

Take the heat to kinetic energy converter. Turn that converter into a cylinder mounted firmly on the ground and a piston that is pushed by steam pressure. If you push it from 0 to 1 m/s, it will not move very far and will not use up much steam. If you push it from 10 to 11 m/s it will move much farther and use up much more steam.

Similarly, if you try are driving your car at 100 kph and brake to a stop, you will cover four times as much distance as if you'd braked from 50 kph to a stop. You will have dissipated four times as much kinetic energy. Your brakes will have heated up four times as hot and you'll have left four times as much rubber on the road.

 

[FactChecker]

Intuition likes to jump immediately to a linear relationship -- but linear to what? As @pixel 's link indicated, linear with respect to distance implies proportional to velocity squared. So you should retrain your intuition to accept "linear with respect to distance". That should be easy. After all, it should take as much energy to lift a weight from 0 ft to 1 ft as it does to lift it from 10,000 ft to 10,001 ft. The jump to velocity squared is simple math.

PS. One nice thing about energy being proportional to v² is that energy does not depend on the sign of velocity, which intuition tells us it should not.

 

[russ_watters]

Some others have provided details of how power and acceleration can be related, which is ok, but I think for this question it is better to recognize first that they aren't intrinsicly related and can only be related by adding-in somewhat arbitrary assumptions about reference frame.

F=ma and a=f/m and w=fd (and p=fd/t) and you can combine them in some cases, but that doesn't mean power and acceleration are automatically related. Heck, in many real world situations, those aren't even the same forces!

Consider a car on a highway at a certain speed. You put the pedal to the floor and the engine produces a certain power. Does it matter if the car is accelerating or if you have your other foot on the brake? It doesn't: the power applied to the drive-shaft by the engine is the same either way.

Perhaps even worse, you can choose to change the power just by changing your frame of reference. This is particularly obvious for a rocket, but is true for a car as well. Say you are following another car on the highway at 50m/s (each has a 2000kg mass). Then you accelerate to 60m/s and run into the back of them. What was your kinetic energy at the time of the collission? It depends on the chosen reference frame, but the reference frame that tells you about the damage in the collission has you moving at 10m/s, not 60m/s.

[net] Force causes acceleration. Power (energy) is derived from other characteristics of the system and may have very little directly to do with the acceleration.