Kinetic energy change with initial velocity

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Discussion Overview

The discussion revolves around the calculation of changes in kinetic energy as an object accelerates from various initial velocities to final velocities. Participants explore the implications of different formulas for kinetic energy and the effects of frame dependence on energy calculations.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents a formula for calculating the change in kinetic energy and questions its application when the initial velocity is not zero.
  • Another participant emphasizes the frame dependence of kinetic energy, suggesting that the energy calculations must consider the relative motion of the object and the target.
  • There is a discussion about whether the same kinetic energy formulas apply when the initial velocity is negative, with participants suggesting that energy considerations may differ based on the context of the collision.
  • Some participants propose that energy can be conserved in elastic collisions, where no net change in kinetic energy occurs despite changes in velocity.

Areas of Agreement / Disagreement

Participants generally agree on the application of the kinetic energy formulas in specific scenarios, but there is no consensus on the implications of negative velocities and frame dependence. The discussion remains unresolved regarding the application of formulas in cases of negative initial velocities.

Contextual Notes

Participants note that the calculations depend on the definitions of kinetic energy and the conditions of the collisions (elastic vs. inelastic), which may not be fully resolved in the discussion.

hipokrytus
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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?
 
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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 ?
 
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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.
 
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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 the initial velocity is 5 m/s, and final is -5 m/s. Would the same formula be used?
 
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.
 
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hipokrytus said:
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'.]
 
Right. Thanks for quick and clear replies.
 

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