Confused about work-energy

  • #1

Main Question or Discussion Point

In my text, we learn that the work done on an object is equal to the change in kinetic energy of that object. It seems to me that this statement requires some restrictions. If I take a an object at rest of mass m and lift to a height h and place it at rest on a shelf, the work I've done is -mgh. But I haven't changed the kinetic energy.
If you consider some U-K system (sorry), perhaps where U is given by mgh, then the work done by the conservative force which provides U is always equal to the negative of the change in kinetic energy, provided there are no other forces.
In the first example, the total energy changes and W=deltaK fails. In the second, the total energy is constant and W=deltaK holds. How do I know in general when to apply the work-energy theorem??
 

Answers and Replies

  • #2
enigma
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Your text isn't giving you the whole story.

The work done on an object = the sum of all other energies. That included Gravitational Potential energy (from your example), thermal energy, electrical energy, kinetic energy, ...

You're definately on the right track since you saw there was something askew.
 
  • #3
Chi Meson
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I might be misunderstanding you, enigma, but what you said sounds a bit off to my classical ear. The work energy theorem states that the total (net) work on an object is equal to the change in the kinetic energy.

When I lift an object to place it on a high shelf, I do +mgh of work on it, but at the same time gravity does -mgh. The net work is zero, so the change in KE is zero. If I were to apply a force greater than the object's weight, then after I lift it a certain height , h, it will be moving with a certain KE, and continue to rise to another height.

When friction is present, it also does negative work. So Wnet=Wapplied-Wgravity-Wfriction

Any other forces that are on the object at the same time will do work on the object according to the relationship W=Fd cos [theta].

Stephen, I am assuming that inclusion of thermodynamics and electrical effects are a tiny bit beyond your current location in the text? Ultimately these must be considered as well, as enigma mentions.

THis work-energy theorem is actually a restatement of the Law of Conservation of energy. Sometimes it's easier to examine problems by comparing initial and final energy states.
 
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  • #4
Originally posted by Chi Meson
The work energy theorem states that the total (net) work on an object is equal to the change in the kinetic energy.
Yes! I always screw that part up. I always forget that it's the net work we're dealing with.
So if a motor elevates an elevator initially at rest 1/2a meters in one second with an acceleration of a meters per second squared, the net work W on the elevator is 1/2ma2 since the velocity changes from 0 to a. The work done by gravity is -mga/2. So the work done by the motor Wm is found by solving W=Wm+Wg for W_m. We find W_m=1/2ma(a+g).

Is it also true that W_net=-[del]U?
Well, -[del]U=mga/2 and W_net=1/2ma2
So g=a? I don't know about that...
What are the conditions regarding W=-[del]U?
 
  • #5
pmb
Originally posted by StephenPrivitera
In my text, we learn that the work done on an object is equal to the change in kinetic energy of that object. It seems to me that this statement requires some restrictions. If I take a an object at rest of mass m and lift to a height h and place it at rest on a shelf, the work I've done is -mgh. But I haven't changed the kinetic energy.
If you consider some U-K system (sorry), perhaps where U is given by mgh, then the work done by the conservative force which provides U is always equal to the negative of the change in kinetic energy, provided there are no other forces.
In the first example, the total energy changes and W=deltaK fails. In the second, the total energy is constant and W=deltaK holds. How do I know in general when to apply the work-energy theorem??
If you're lifting the box at a constant rate then the net force is zero so that no work is being done on the object.

Pete
 
  • #6
Chi Meson
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>>Is it also true that W_net=-U?
Well, -U=mga/2 and W_net=1/2ma2
So g=a? I don't know about that...
What are the conditions regarding W=-U?<<

No.

I really have a hard time with the subscripts in this format, so I'll use the style you're using. Please excuse in advance my style if it seems pedantic.

The work-energy theorem states W_net = delta K .

In translation, W_net is the sum of each of the works done by all forces on an object during a time interval. Let's say all the works from non-perpendicular forces are: from an applied force (some guy pushing), W_p ; friction, -W_f ; gravity, -W_g.

If a cart is pushed up a ramp, and it had an initial velocity, it would have an initial and a final KE, K_o and K_f. It would have an initial and final PE, U_o and U_f.

Compare the initial and final energy states:
K_o + U_o + W_p = K_f + U_f + Q

W_net = W_p - W_f - W_g
W_g = delta U = U_f - U_o
Friction turns kinetic energy into thermal energy, so -W_f = -Q
delta K = K_f - K_o

Substitute and manipulate, you get

W_net = delta K
 

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