Energy conservative (Kinetic Energy and Potential Energy)

AI Thread Summary
The discussion centers on the confusion surrounding the conservation of energy in a scenario where a book is lifted at constant velocity. It highlights that while kinetic energy (K) remains constant (K2 - K1 = 0), potential energy (U) changes (U1 ≠ U2) due to the work done by an external force. The equation for work done by non-conservative forces is clarified as W_nc = ΔU + ΔK, emphasizing that mechanical energy is not conserved when non-conservative forces are involved. Participants stress the importance of distinguishing between work done, kinetic energy, and potential energy in such systems. Ultimately, the conversation reinforces that in isolated systems with only conservative forces, total mechanical energy remains constant.
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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