When I moving object in some frame of reference, it

spazzter08dyk2n

spazzter08dyk2n

Answered question

2022-05-18

When I moving object in some frame of reference, it has a positive kinetic energy. In the frame of reference where the object is stationary, it has a kinetic energy of 0.
Is the energy difference in some way associated to the frames of reference?

Answer & Explanation

Percyaehyq

Percyaehyq

Beginner2022-05-19Added 18 answers

"Does a frame of reference have energy in itself?"
No, there's no reason to think of it that way. When we talk about energy, it's very useful because it can be transferred between object and move from one form to another.
There's no transformation we can do that will allow us to extract energy from a frame. So the energy differences in your scenario aren't because the energy is "in the frame".
We can imagine a frame where you are moving at 0.5 c. Your body has a numerically large value for kinetic energy in that frame. But you have no means to exploit that energy.
Rather than the absolute value of K E, you should think about how energy moves from one thing to another. Take a pair of objects and have them collide inelastically. Some of the K E is turned into thermal energy. In the frame where the center of mass is at rest, all the K E is used in this manner. In a frame where the center of mass is at high speed, K E moves from one object to the other, but the total loss of kinetic energy into heat is identical. In neither case do we need to consider the frame to be involved in the energy transfer.
"Does that mean a body that moves somehow needs a collision for the kinetic energy to manifest?"
A "collision" may not be necessary, but an "interaction" with another mass is. You can't change K E of an object without changing velocity, and you can't change velocity without changing momentum, and momentum is conserved. If there's no interaction, then there's no change in K E. Without being able to change it, the numerical value has no meaning.
A large K E is only meaningful in scenarios where the K E can be reduced (giving energy to something as it slows down).
It accelerates due to gravitational interaction with the earth. You may not be able to measure the acceleration of the earth in this interaction, but it is present. The potential energy in the gravitational system of the ball and the earth, and the kinetic energy of the earth all change as the ball falls. After you let go of the ball, the total of all three is constant (in any inertial frame you choose). Different frames would have different absolute values, but the energy transfer always sums up.
E t o t = K E b a l l + K E e a r t h + G P E e a r t h b a l l s y s t e m
Nubydayclellaumvcd

Nubydayclellaumvcd

Beginner2022-05-20Added 5 answers

Yes it is. The macroscopic kinetic energy of an object (the kinetic energy of an object as a whole) is its kinetic energy with respect to an external (to itself) frame of reference.
For example, a car of mass m moving with velocity v relative to a person standing on the road, has a kinetic energy of m v 2 2 with respect to the person on the road. On the other hand, its kinetic energy with respect to a person in another car driving next to it in the same direction with the same velocity with respect to the road is zero, because the velocity of the first car with respect to the second car is zero.

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