Kinetic energy change with initial velocity

  • #1
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:

Answers and Replies

  • #2
BvU
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Hi hipo,

What energy would you need to change an intital velocity of 0 m/s to 40 m/s in one go ?
 
  • #3
sophiecentaur
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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.
 
  • #4
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?
 
  • #5
BvU
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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.
 
  • #6
sophiecentaur
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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 me initial velocity is 5 m/s, then -5 m/s. Would the same formula be used?
How could it possibly not apply - as long as you use the right sums? If you do it in two steps, you can get energy out of slowing down and return it - so no net KE change in the Earth frame (you could use a steel spring, mounted on a fixed base). Best to use -10 and +10 m/s for comparison actually, I think. Exactly the same situation for a source and target moving at -10m/s (relative to Earth)
[Edit BvU got there first about the 'bounce'.]
 
  • #7
Right. Thanks for quick and clear replies.
 

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