# Conservation of Energy

estro
Suppose that I hold a Yo-Yo and then I leave it to fall [and of course rotate] holding only the string.
I seems to me that the energy [not the whole of the potential energy will be converted to $(1/2)mv^2+(1/2)I\omega^2$] is not conserved, because of the force the string applies on the Yo-Yo.
Is my intuition is right?

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Mentor
Suppose that I hold a Yo-Yo and then I leave it to fall [and of course rotate] holding only the string.
I seems to me that the energy [not the whole of the potential energy will be converted to $(1/2)mv^2+(1/2)I\omega^2$] is not conserved, because of the force the string applies on the Yo-Yo.
Is my intuition is right?
I would say no. Just because the string exerts a force on the yo-yo does not mean that work is being done. Consider the speed of the end of the string as it unravels.

estro
I would say no. Just because the string exerts a force on the yo-yo does not mean that work is being done. Consider the speed of the end of the string as it unravels.

Regarding what you said about the work:
The yo-yo goes down so the string force does negative work. So the force does not "spend" energy, right?

Regarding speed of the string:
The string has no speed after it unravels, but I don't understand how is this related to work. As work is only $F\centerdot \Delta x$

Mentor
Regarding what you said about the work:
The yo-yo goes down so the string force does negative work. So the force does not "spend" energy, right?

Regarding speed of the string:
The string has no speed after it unravels, but I don't understand how is this related to work. As work is only $F\centerdot \Delta x$
Work is better thought of as $F\centerdot \Delta x$, where $\Delta x$ is the displacement of the point of application of that force. But the point of application is not moving.

You can certainly calculate $F_{net}\centerdot \Delta x_{cm}$, but that's not the same thing as the work done by the forces.

estro
Work is better thought of as $F\centerdot \Delta x$, where $\Delta x$ is the displacement of the point of application of that force. But the point of application is not moving.

You can certainly calculate $F_{net}\centerdot \Delta x_{cm}$, but that's not the same thing as the work done by the forces.

Great explanation [cleared a lot of confusion for me], thanks!