# Kinetic energy change with initial velocity

It is my understanding that to calculate the change of kinetic energy of an object that speeds up from vi to vf you use this formula:
Change of kinetic energy = 1/2 * m * (vf2 - vi2)

When the initial velocity is 0 m/s I have no problems, but let's say an object that weighs 2 kg speeds up from 20 m/s to 40 m/s.
When I use the formula I mentioned, I get:
ΔEk = 1/2 * 2 * (1600 - 400) = 1200 J

Now let's say the initial velocity is 0 m/s, final 20 m/s. I then get:
ΔEk = 1/2 * 2 * (400 - 0) = 400 J

Does this make sense? The same changes in velocity and different changes in kinetic energy?
Should I not use a formula like this instead:
ΔEk = 1/2 * m * (Δv)2 = 1/2 * m * (vf - vi)2
?
Then I end up with 400 J in both cases.

Also, what if the initial velocity is, lets's say 5 m/s and final -5 m/s? Does the kinetic energy change?

Last edited:

BvU
Homework Helper
Hi hipo,

What energy would you need to change an intital velocity of 0 m/s to 40 m/s in one go ?

hipokrytus
sophiecentaur
Gold Member
KE is frame dependent, so you have to be careful to compare the right things. If you look upon it as the damage that a projectile could inflict on a stationary (= Earth frame) target, with an initial velocity of 20m/s you already have 400J and the extra 20m/s gives a relative speed of 40m/s. The 'extra' energy is there because the Work to provide the speed increase would be Force times Distance. The distance is much more when you start at 20m/s than when you start at 0m/s.
Why not do the sums and prove it for yourself? It would be a good exercise.
PS. The damage that your projectile from a 20m/s start would be only 400J worth if it impacted on a target also going at 20m/s. That's how the frame dependence comes in.

hipokrytus
Hi,
Ok, I get it. From 0 m/s to 20 m/s it's 400 J, from 20 m/s to 40 m/s it is 1200 J, it adds up to 1600 J, which is the same as from 0 m/s to 40 m/s. So the 1st formula works. But what about the situation when the initial velocity is 5 m/s, and final is -5 m/s. Would the same formula be used?

BvU
Homework Helper
If you use e.g. a rubber wall, no energy is needed to change a speed of 5 m/s to -5 m/s. Condition is that the collision process is elastic.

hipokrytus
sophiecentaur