Energy conservative (Kinetic Energy and Potential Energy)

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Homework Help Overview

The discussion revolves around the conservation of energy in the context of lifting a book with a constant velocity. Participants explore the relationships between work done, kinetic energy, and potential energy, questioning how these concepts apply when external forces are involved.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the implications of lifting a book at constant velocity, noting that kinetic energy changes may not occur while potential energy does. They question the validity of energy conservation equations in the presence of non-conservative forces.

Discussion Status

There is an ongoing exploration of the relationships between work, kinetic energy, and potential energy, particularly in systems with both conservative and non-conservative forces. Some participants suggest specific equations while others express confusion about the underlying principles.

Contextual Notes

Participants highlight the need to consider external forces, such as friction and the lifting force, when analyzing energy changes. There is also mention of the distinction between isolated systems and those influenced by external forces.

MechaMZ
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Homework Statement


May i know how do i compute an energy conservative equation for the scenario below, which i don't understand.

If a book is lifted up by a force in order to make it moving upward with a constant velocity.
For conservation of total energy, W1-2 = U1-U2 = K2 - K1
If the object is moving in a constant velocity, isn't K2 - K1 = 0
However, U1 is not equal to U2.

It seems not correct obviously, hope someone could enlighten me.

I guess W1-2 = U1-U2 = K2 - K1 is only valid for isolated system, which should be a system without any external force involves except gravitational force.

I'm very confuse now, hope someone could help me.
 
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w=f.d
f=m.g
d=distance moved
earth does -ve work , you do equal +ve.


ok!

+ve change in potiential energy always needs external energy source.

so W=U2-U1=m.g.d
 
MechaMZ said:

Homework Statement


May i know how do i compute an energy conservative equation for the scenario below, which i don't understand.

If a book is lifted up by a force in order to make it moving upward with a constant velocity.
For conservation of total energy, W1-2 = U1-U2 = K2 - K1
If the object is moving in a constant velocity, isn't K2 - K1 = 0
However, U1 is not equal to U2.

It seems not correct obviously, hope someone could enlighten me.

The work done by non conservative forces (such as the lifting force on the book) is
W_{nc} = (U_2 - U_1) + (K_2 - K_1) or in its alternate form, W_{nc} = \Delta U + \Delta K

I guess W1-2 = U1-U2 = K2 - K1 is only valid for isolated system, which should be a system without any external force involves except gravitational force.

I'm very confuse now, hope someone could help me.
For a system subject to conservative forces only (like gravity or spring forces), then (U_2 - U_1) + (K_2 -K_1) = 0, or \Delta U + \Delta K =0, and the work done by such conservative forces is W_c = U_1- U_2, or W_c = - (\Delta U) in its alternate form.

For a system subjct to both conservative and non conservative forces, then the total work done by those forces is W_T = (K_2- K_1), or W_T= \Delta K

staraet's answer for the work done by the lifting force is also correct.

This can get confusing for sure, but actually, I'm not exactly sure what your question is. Mechanical Energy (U + K) is NOT conserved when non-conservative forces that do work are acting.
 
Last edited:
PhanthomJay said:
For a system subjct to both conservative and non conservative forces, then the total work done by those forces is W_T = (K_2- K_1), or W_T= \Delta K

why we no need to include the \Delta U? What if the object is in a falling motion with air resistive force acting on it.

Could i write it as: mg(\Delta h) - Ffriction(\Delta h) = \Delta K + \Delta U

or i should write it as without taking the work done by gravitational force: - Ffriction(\Delta h) = \Delta K +\Delta U
 
always think in terms of gain and loss

we some thing goes along only acting force , it gains k.e
so decreases its p.e

when ball falls, so it means, going along a force, so increases k.e (say after dt);
then if u find k.e is constant-- it means body has spent that k.e in either heat, vibrational,mass or something in that dt time.
then write equations. don't write before....

here friction is heat. so E spent in heat= mgh
 
I think my problem is I don't understand between work done, kinetic energy, and potential energy.
 
I suggest sticking with PhanthomJay's first equation:
PhanthomJay said:
The work done by non conservative forces (such as the lifting force on the book) is
W_{nc} = (U_2 - U_1) + (K_2 - K_1) or in its alternate form, W_{nc} = \Delta U + \Delta K.

As you realize for your example of lifting a book, K2=K1=0. Also, Wnc is the work W1-2 done by lifting, so in that case

W1-2 = U2 - U1

If instead we have an isolated system with only conservative forces such as gravity, we have Wnc = 0 and conservation of mechanical energy applies. PhanthomJay's equation then becomes:

0 = U2 - U1 + K2 - K1

this is often rewritten in the form

U1 + K1 = U2 + K2

and is simply a statement that the total energy (potential + kinetic) does not change.
 

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