# Kinetic energy change with initial velocity

• hipokrytus
In summary, when objects speeds up, the change of kinetic energy is calculated using this formula: Change of kinetic energy = 1/2 * m * (vf2 - vi2)
hipokrytus
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:
Hi hipo,

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

hipokrytus
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?

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
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.

## 1. What is kinetic energy and how is it related to initial velocity?

Kinetic energy is the energy that an object possesses due to its motion. It is directly proportional to the square of the object's initial velocity, meaning that as the initial velocity increases, the kinetic energy also increases.

## 2. How does the mass of an object affect the change in kinetic energy with initial velocity?

The mass of an object does not directly affect the change in kinetic energy with initial velocity. However, since kinetic energy is directly proportional to the square of the initial velocity, a heavier object will require more initial velocity to have the same change in kinetic energy as a lighter object.

## 3. What is the formula for calculating the change in kinetic energy with initial velocity?

The formula for calculating the change in kinetic energy with initial velocity is KE = 1/2 * m * (v2 - u2), where KE is the change in kinetic energy, m is the mass of the object, v is the final velocity, and u is the initial velocity.

## 4. Can the change in kinetic energy be negative with an increase in initial velocity?

No, the change in kinetic energy cannot be negative as it is always a positive value. However, if the initial velocity is greater than the final velocity, the change in kinetic energy will be negative, indicating a loss of kinetic energy.

## 5. How does friction affect the change in kinetic energy with initial velocity?

Friction can decrease the initial velocity of an object, resulting in a smaller change in kinetic energy. This is because friction converts some of the initial kinetic energy into other forms of energy, such as heat or sound.

Replies
3
Views
505
Replies
1
Views
946
Replies
13
Views
1K
Replies
3
Views
1K
Replies
10
Views
1K
Replies
5
Views
1K
Replies
9
Views
2K
Replies
2
Views
1K
Replies
16
Views
1K
Replies
11
Views
1K
The text passage contains a discussion about the calculation of kinetic energy changes with different initial and final velocities. The user poses several questions and provides some calculations, while another user responds with explanations and further questions.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Kinetic energy change with initial velocity

• hipokrytus
In summary, when objects speeds up, the change of kinetic energy is calculated using this formula: Change of kinetic energy = 1/2 * m * (vf2 - vi2)
hipokrytus
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:
Hi hipo,

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

hipokrytus
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?

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
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.

## 1. What is kinetic energy and how is it related to initial velocity?

Kinetic energy is the energy that an object possesses due to its motion. It is directly proportional to the square of the object's initial velocity, meaning that as the initial velocity increases, the kinetic energy also increases.

## 2. How does the mass of an object affect the change in kinetic energy with initial velocity?

The mass of an object does not directly affect the change in kinetic energy with initial velocity. However, since kinetic energy is directly proportional to the square of the initial velocity, a heavier object will require more initial velocity to have the same change in kinetic energy as a lighter object.

## 3. What is the formula for calculating the change in kinetic energy with initial velocity?

The formula for calculating the change in kinetic energy with initial velocity is KE = 1/2 * m * (v2 - u2), where KE is the change in kinetic energy, m is the mass of the object, v is the final velocity, and u is the initial velocity.

## 4. Can the change in kinetic energy be negative with an increase in initial velocity?

No, the change in kinetic energy cannot be negative as it is always a positive value. However, if the initial velocity is greater than the final velocity, the change in kinetic energy will be negative, indicating a loss of kinetic energy.

## 5. How does friction affect the change in kinetic energy with initial velocity?

Friction can decrease the initial velocity of an object, resulting in a smaller change in kinetic energy. This is because friction converts some of the initial kinetic energy into other forms of energy, such as heat or sound.

Replies
3
Views
505
Replies
1
Views
946
Replies
13
Views
1K
Replies
3
Views
1K
Replies
10
Views
1K
Replies
5
Views
1K
Replies
9
Views
2K
Replies
2
Views
1K
Replies
16
Views
1K
Replies
11
Views
1K
user
The text passage contains a discussion about the calculation of kinetic energy changes with different initial and final velocities. The user poses several questions and provides some calculations, while another user responds with explanations and further questions.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS