# Why is velocity squared in kinetic energy? [duplicate]

I'm not looking for how it's derrived. I'm wondering why it works out this way that doubling the speed quadrouples the KE.

Here's my thought process:

A car is moving down the road (ignoring air resistance and any other external force reliant on speed) at 10 m/s. To get to 20 m/s the engine should burn x amount of gas, containing x amount of energy. To do the same acceleration over again it should burn x gas again, resulting in 2x gas burned total and resulting in another 10 m/s increase in speed. This would leave it at 30 m/s.

Say the amount of gas burned to increase speed by 10 m/s is 100 energy units, and the car's mass is 2. Then increasing from 10 to 20 with ke = 1/2 mv2 should increase the KE by 100. Therefore from 20 to 30 it should also be 100, resulting in 200 total energy units and 2x gas burned. But using the equation and plugging in the same mass value with the total acceleration from 10 to 30 (30 -- 10 = 20 and plugging that in for v rather than 10) you get the total ke as 400 units rather than 200, but you theoretically should have still used 2x gas.

If someone could explain to me what the issue is in my reasoning and explain why speed is squared, that would be much appreciated.

• Because work is cumulative. It is an integral equation. Jan 11, 2018 at 17:11
• Jan 11, 2018 at 18:33
• Honestly, Infeel the best way to know this is by drawing a VT graph Jan 11, 2018 at 21:47

Imagine you're towing the car by pulling a rope with your hands. If you always pull with the same force, then the car always has the same acceleration. Then it takes just as much time to get the car from 10 mph to 20 mph as it does to take it from 20 mph to 30 mph, as you said.

But that doesn't mean it takes the same amount of energy. You'll be working much harder in the second half, because you'll have to pull in the rope faster. That's more strenuous. It's like switching from doing a pullup every two seconds to a pullup every second.

The same reasoning holds for the motor; it has to work harder to exert a force on something that's already moving fast.

• You'll be working much harder in the second half, because you'll have to pull in the rope faster. Perhaps more obvious to say that you do more work because you apply the force over a longer distance. A force applied for 1 min at an average of 25 mph is applied over a longer distance than when applied for 1 min at an average of 15 mph. Jan 11, 2018 at 18:27
• @sammygerbil True, but I’m just trying to give intuition for why it should “feel” harder, without having to use the definition of work. Jan 11, 2018 at 18:28
• Why would energy be proportional to the velocity? because you wish so? Jan 11, 2018 at 18:48
• @ŽarkoTomičić It's just to give intuition. Of course the real answer is that the word "energy" is just defined to make it so. Jun 28, 2019 at 12:44

Your error lies in assuming there's a fixed amount of energy (100 units) for 10mph of velocity increase. There's no basis for that assumption.

The acceleration from 20 to 30 is harder than 10 to 20 because the ground you're pushing against with the wheels is already going 20mph against you instead of just 10.

It's like when you push on a skateboard - initially it's easy to accelerate, but as you go faster you have to get your foot going faster than ground speed to get even any acceleration. Hence energy is proportional to velocity squared.

• Also, if x is the amount of fuel for some energy gain and if you go from 10 to 20 mph than you just quadrupled the energy that you had....and now, it means that x amounts to 3 times the energy you had. But from 20 to thirty, speed is 1.5 times higher and you have just increased the energy you had 2.25 times.... Jan 11, 2018 at 18:32