Need help understanding work equations :-)

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In summary: When you are using the conservation of energy, it is best to not remember a formula but just remember what is the conservation of energy. The total energy in a closed system is constant. Closed system meaning that you aren't supplying external "energy" to the system.Now if a particle is moving up the plane and a constant force,F, is pushing it up the plane a displacement,x, the work done by the force is,Fx.But as the block moves up the plane, it has a speed right? So, by the kinetic energy formula (\frac{1}{2}mv^2) it has kinetic energy,E_k.Also as the
  • #1
TNCEKM
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Okay, so I'm a biology major who has already finished life-sciences physics, but I'm reviewing the material right now for the MCAT and the one subject that has got me a little confused is work. The texts I have available to me are essentially giving me equations and telling me to memorize them, but I find it much more useful to be able to understand the equations and what they mean so I can derive other equations from them.

Anyway, does anybody have a link available to them that will help explain the basic work equations?

Thanks!
 
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  • #2
Which equations exactly do you want to know how to derive?
 
  • #3
Well, I really need help understanding what the more general equations stand for and how they work so I can derive more specific equations from them. Where I really get confused is when work is thrown into a conservation of energy statement--it screws with my previous knowledge.

Given I'm superficially familiar with the following equations:
1) W = ΔU + ΔK + ΔE
2) W = ΔU + ΔK = ΔE
3) W = ΔK (Work-Energy theorem)
4) W = -ΔU

* 1 and 2 both being correct equations just doesn't make sense to me.
* I would expect to be able to derive W = -ΔU from Either (1) or (2), but the sign convention screws me up and in the case of something like a moving block down an inclined plane, there is both a ΔK and a ΔU, so I can't isolate one or the other to give me (4).

I just find it all confusing. Maybe there are some distinctions between variables I'm not picking up, but this stuff just doesn't make sense to me as it was presented to me in the equations I listed above.

Sorry if I'm a bit scatter-brained in my questioning... I just don't see the connections here!
 
  • #4
1 and 2 would make sense if E in case one was like a "thermal energy" and E in case 2 was a "total energy"... is this the case? Case 1 would just be when no heat is involved and case 2 when no heat or friction?

But, that still leaves me confused with how (3) and (4) related to (1) and (2)

Thanks!
 
  • #5
When you are using the conservation of energy, it is best to not remember a formula but just remember what is the conservation of energy. The total energy in a closed system is constant. Closed system meaning that you aren't supplying external "energy" to the system.

Now if a particle is moving up the plane and a constant force,F, is pushing it up the plane a displacement,x, the work done by the force is,Fx.

But as the block moves up the plane, it has a speed right? So, by the kinetic energy formula ([itex]\frac{1}{2}mv^2[/itex]) it has kinetic energy,[itex]E_k[/itex].

Also as the block moves up the plane it moves further away from the ground,correct? (As it is basically moving upwards in a sense). So it must be gaining gravitational p.e.,[itex]E_p[/itex]

Now if the plane is rough (i.e. frictional forces apply), work is done overcoming that friction,else the block just wouldn't move. The work done in overcoming friction is E. (You calculate it by finding the product of the frictional force and the distance moved)

Since energy can't be created or destroyed, the energy must have been converted from something. What is this something? Well it is the initial work done by the force,Fx!

So that Energy input to the system= Energy output (by the law of conservation of energy)

and so

[itex]Fx=E_k+E_p+E[/itex] ...(that's equation 1)

and that is basically how to use it.

If the plane were such that you can ignore friction, well then that means that E=0 and hence

[itex]Fx=E_k+E_p[/itex] ...(that's equation 2)

(ignoring friction from here now)
If a box is pushed along a surface by a Force then the Force does work,W, in pushing the box. Since you are ignoring friction and the height of the box didn't change, then all the work done by the force is converted to kinetic energy which is used to give the box a speed.
So that

[itex]W=E_k[/itex]...(That's equation 3)
 
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  • #6
Thanks for the response.

So, I've got a good handle on PE and KE themselves, and how PE is related to position and KE is related to relative motion. And, now equations 1 and 2 make more sense given that E from equation 1 is different than E from equation 2. Equation 3 is good, too.

But, I still have a few more questions, if you don't mind :)

A) Just to clarify, Equation 1 = Equation 2, but equation 2 would be in the case where there is no friction. So does that mean I could restate equation 1 as: W = ΔU + ΔK + ΔEi = ΔE?

B) How does equation 1 relate to equation (W = -ΔU)

C) In the case friction is present, is mechanical energy still conserved?

Thanks!
 
  • #7
TNCEKM said:
A) Just to clarify, Equation 1 = Equation 2, but equation 2 would be in the case where there is no friction. So does that mean I could restate equation 1 as: W = ΔU + ΔK + ΔEi = ΔE?
Nope. You can't because in equation one and two there are different forces acting, so they are different systems.


TNCEKM said:
B) How does equation 1 relate to equation (W = -ΔU)
Consider a body of mass,m, at some height above the ground and moving upwards with an arbitrary velocity. If the body is to maintain this velocity,there must be an upward force acting on it equal to its weight,mg. If the body moves upwards by a further distance,h, under the influence of this same force (which is constant), the work done,W is given by.
W=(mg)*h {mg is a Force and h is the distance moved}
=mgh.
There can have been no change in k.e. since the velocity was the same and therefore the work done by the body was only used to increase the potential energy.


TNCEKM said:
C) In the case friction is present, is mechanical energy still conserved?

Yes, mechanical energy is conserved. If friction is present, then all it means is that some work is done in overcoming friction.
 
  • #8
Thanks, I appreciate it!

(Man, of all the areas for a person to get hung up! One would assume it would have been E&M or fluids, or something... but nooooo... for me its "work" *sigh*)
 
  • #9
TNCEKM said:
Thanks, I appreciate it!

(Man, of all the areas for a person to get hung up! One would assume it would have been E&M or fluids, or something... but nooooo... for me its "work" *sigh*)

Well it helps if the textbook you use gives you the equations and explains how to arrive at them.

But we all get confused about some topic at one time or the other...I mean I got confused with errors at one point in time!
 
  • #10
rock.freak667 said:
Well it helps if the textbook you use gives you the equations and explains how to arrive at them.

But we all get confused about some topic at one time or the other...I mean I got confused with errors at one point in time!
I guess that's true :) Thanks again for your help!
 

1. What is the definition of work in terms of physics?

In physics, work is defined as the product of force and displacement. It is a measure of how much force is applied over a certain distance.

2. How is work calculated?

The mathematical equation for calculating work is W = F x d, where W is work, F is force, and d is displacement. Work is measured in joules (J).

3. What is the difference between work and power?

Work is the amount of energy transferred when a force is applied over a distance, while power is the rate at which work is done. Power is calculated by dividing work by time, and is measured in watts (W).

4. Can negative work be done?

Yes, negative work can be done. Negative work occurs when the force and displacement are in opposite directions, resulting in a decrease in energy.

5. How is work related to kinetic energy?

According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. This means that work can either increase or decrease an object's kinetic energy.

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