Kinetic energy and potential energy concept question

[channel1]

i need a review. why does lifting a book at a constant speed result in a constant KE but changing PE? obviously KE = 1/2 mv^2 but I am looking for a conceptual answer. doesn't KE need to equal PE for energy to be conserved?

im in e&m right now and there's a statement in my book saying that the book situation is analogous to potential within an emf source. so W = 0 (work done on a test charge within an ideal emf source), U is increased, and KE is constant. but in the book analogy doesn't the book have to be moving in the +/- x direction for W = 0?

so to summarize i have two questions, (1) concerns how energy is conserved with a constant KE but changing U and (2) concerns how W = 0 if there is a changing U

please answer both questions in terms of the book analogy AND the emf source! thanks in advance :)

 

[NascentOxygen]

i need a review. why does lifting a book at a constant speed result in a constant KE but changing PE? obviously KE = 1/2 mv^2 but I am looking for a conceptual answer.

Lifting a book is a quite different situation to a falling book. In lifting it, you are doing work all the while, so you are increasing its potential energy by moving it higher.

doesn't KE need to equal PE for energy to be conserved?

For something falling, its PE is converted into KE. No extra work is being done on the system. One form of energy is being converted into the other. And that KE all gets wasted when the book hits the floor.

 

[channel1]

that makes sense but how is there zero work being done in an emf source yet there is a changing KE and static PE? :/ its almost like the falling book situation and lifting book situation combined...

 

[channel1]

strike that, reverse it :) changing PE, static KE

 

[rwisz]

First of all, it is important to understand the difference between kinetic energy (KE) and potential energy (PE). KE is the energy an object possesses due to its motion, while PE is the energy an object possesses due to its position or configuration.

In the case of lifting a book at a constant speed, the KE remains constant because the book is not accelerating, meaning its velocity (v) remains constant. This is reflected in the formula KE = 1/2 mv^2, where v is squared. This means that even if the book is lifted to a higher position, the increase in potential energy does not affect the kinetic energy since there is no change in velocity.

On the other hand, potential energy changes because the book is being lifted against the force of gravity, which is constantly pulling the book downwards. As the book is lifted to a higher position, it gains more potential energy because it has the ability to do more work when released. This increase in potential energy is equal to the work done against gravity (W = mgh).

Now, to answer your first question, energy is still conserved even though the KE and PE are changing. This is because when the book is lifted, the work done against gravity is stored as potential energy. When the book is released, the potential energy is converted back into kinetic energy as the book falls. The total energy (KE + PE) remains constant throughout this process, as energy cannot be created or destroyed.

In terms of the book analogy and the emf source, we can think of the book as the test charge and the force of gravity as the emf source. The work done against gravity is analogous to the work done on a test charge within an ideal emf source. In this case, the work done (W) is equal to the change in potential energy (U) because there is no change in kinetic energy.

For your second question, in the book analogy, the book does not have to be moving in the +/- x direction for the work done to be zero. As long as the book is moving at a constant speed, the work done against gravity is equal to the change in potential energy, which is zero since there is no change in height. Similarly, in the emf source, the work done on the test charge is zero because the test charge is moving at a constant speed and there is no change in potential energy.

In summary, lifting a book at a constant speed