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I posted a similar question yesterday, but the question hasn't been solved yet, so I'm posting a similar question again. (sorry for similar question..)

Let me write my interpretation of the situation below.

First case) On frictionless ice, 50kg of people exert force on 200kg of objects. (Initial speed is zero.)

If the later speed of an object weighing 200 kg is 1 m/s, a person weighing 50 kg will have a speed of 4 m/s in the opposite direction.

The kinetic energy of an object is 100 J and the kinetic energy of a person is 400 J, and the total kinetic energy is 500 J. This means that a person has consumed 500 J of energy.

This is true even if a person does 100J of work for an object, and an object does 400J of work for a person.

Second case) It's a similar situation, but let's think about a slightly different situation. A 200-kilogram object is on frictionless ice. Let's say that someone pushes this object by 1 meter with a force of 100 N.

And what a human does to an object is 100 J as above. The later speed of the object will be 1 m/s as above.

The object will also push a person by 1 meter with a force of -100 N (reaction force), so what the object has done to a person is -100 J.

Something's wrong.

What a person does to an object is equal to 100J in the first case and in the second case. But in the first case, the total kinetic energy is 500 J, which is about as much as 400 J.

Where did the error come from?

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  • $\begingroup$ In the second case are both person and object on the ice? $\endgroup$
    – Farcher
    Commented Apr 18, 2023 at 7:17
  • $\begingroup$ @Farcher It's the same situation as situation one. Both humans and objects are on frictionless ice. It's just a difference in interpretation. And I'm wondering what happens if only objects are on frictionless ice and a person is not moving on the ground and only applying force. $\endgroup$
    – NOH WHIREA
    Commented Apr 18, 2023 at 8:37
  • $\begingroup$ @NOHWHIREA said “I posted a similar question yesterday, but the question hasn't been solved yet”. This is flat out a lie. You had 3 answers to that question at the time you wrote this one. I have downvoted both questions, that one because it was poorly written and this one for the lie. Do not write duplicate questions and do not lie about things. physics.stackexchange.com/questions/760068/… $\endgroup$
    – Dale
    Commented Apr 18, 2023 at 11:49
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    $\begingroup$ @Dale, that is also what I remembered. All of those questions are ill-posed because he is just confused about the subject. I only answered because this is slightly better posed. He probably did not even undersand that he asked a slightly different variation. $\endgroup$
    – naturallyInconsistent
    Commented Apr 19, 2023 at 2:00
  • $\begingroup$ You might be interested in physics.stackexchange.com/questions/760273/…, which is inspired by your question. (+1 for the interesting gedankenexperiment) $\endgroup$
    – Allure
    Commented Apr 19, 2023 at 6:47

2 Answers 2

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You are just wrong. Sloppy in the analysis of the issues. And if you have already asked the question before, do not make new ones.

In the first case, the person converted 500J of chemical potential energy (stored food) from muscles and gave 100J of that as kinetic energy for the object, 400J leftover is the kinetic energy of the person. Conservation of energy is perfectly fine with this.

In the second case, the person converted only 100J of chemical potential energy from muscles and gave all 100J of that as kinetic energy for the object. Again, conservation of energy is obeyed.

If you have ever played in an ice skating rink you will know that it is far more difficult to push in the first case than in the second case. There is no paradox at all. The forces, impulses, distances, energies, will have a sensible resolution.

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I made some calculations... In the first scenario, it's already said that the object is moving at 1 m/s and the person is moving at 4 m/s. The sum of their kinetic energy is 500 J, which could only have been provided by the person.

For the second scenario, I assumed one thing: The 200 kg object is moved 1 m with respect to a motionless point, not until it is at 1 m from the person (the person will move as well with respect to the motionless point because of the reaction force).

The 200 kg object is accelerated with a constant force of 100 N. In other words, accelerated at 0.5$\frac{m}{s^2}$. We can find how much time is required to move the object 1 m with this acceleration: $$d = \frac{1}{2}at^2$$ $$t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2(1m)}{0.5m/s^2}} = 2s$$

If the object is accelerated during 2s, it means that the 50 kg person is accelerated as well for two seconds. We find the person's acceleration by dividing the force applied to them by their mass: $$a_p = \frac{100N}{50kg} = 2 m/s^2$$

Now that we know the magnitude of the person's acceleration and for how much time this occurs, we may find their final speed: $$v_f = at = 2\frac{m}{s^2}(2s) = 4m/s$$

By using the kinetic energy formula, we get for the person: $${E_k}_{person} = \frac{1}{2}(50kg)(4m/s^2)^2 = 400J$$

Therefore, energy is conserved. If my assumption wasn't the expected one, energy would still be conserved (of course) but it would be a different scenario than the first one. The object and person would be accelerated for less time and the total kinetic energy would be less as well.

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