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Need help understanding work equations :) 
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#1
Apr208, 05:04 PM

P: 6

Okay, so I'm a biology major who has already finished lifesciences 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! 


#2
Apr208, 05:34 PM

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P: 6,204

Which equations exactly do you want to know how to derive?



#3
Apr208, 05:56 PM

P: 6

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 statementit 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 (WorkEnergy 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 scatterbrained in my questioning... I just don't see the connections here! 


#4
Apr208, 06:08 PM

P: 6

Need help understanding work equations :)
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
Apr208, 06:12 PM

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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) 


#6
Apr208, 07:10 PM

P: 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
Apr208, 07:52 PM

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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. 


#8
Apr208, 07:59 PM

P: 6

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
Apr208, 08:03 PM

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P: 6,204

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
Apr208, 11:02 PM

P: 6




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