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The conservation of energy, why does it work? Negative work vs positive work 
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#1
Dec910, 07:02 PM

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1. The problem statement, all variables and given/known data
This is the conservation of energy in the absence of external forces ΔKE + ΔPE + ΔTE + ΔU= 0 TE is friction, U is other internal energy, like the spring So expand and disregard U and TE for a moment ½m(v²  v₀²) + mg(h  h₀) = 0 Now comes the question, why is + mg(h  h₀) and not mg(h  h₀)? Is gravitational potential energy not the same thing as the work done by gravity? Because gravity pulls down and if you go up, that's negative work. I've used the equation (with the positive mgh) many times and have not failed me. 


#2
Dec910, 09:53 PM

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½m(v²) = mgh, solve for v, and note the change in potential energy is negative, and the work done by gravity is positive. If, on the other hand, an object is thrown upward from the ground at an initial speed v, to a maximum height of h, then ½m(0  v²)+ mg(h  0) = 0, or  ½m(v²) = mgh, solve for v, and note the change in potential energy is positive, and the work done by gravity is negative. Gotta love those plus and minus signs...one of the more difficult concepts of Physics to grasp. 


#3
Dec1010, 01:43 PM

P: 2,568

½m(v²  v₀²) + mg(h  h₀) = 0 <==== this one works or ½m(v²  v₀²)  mg(h  h₀) = 0 <==== this one does not, but it is correct 


#4
Dec1010, 01:56 PM

P: 137

The conservation of energy, why does it work? Negative work vs positive work
Remember that g is 9.81m/s^{2}. Thus, when you plug it into your equation, positive mgh has a negative value.



#5
Dec1010, 02:10 PM

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#6
Dec1010, 02:12 PM

Mentor
P: 41,325




#7
Dec1010, 02:14 PM

P: 137

Oh, right. Oops... Sorry about that.



#8
Dec1010, 02:31 PM

P: 2,568

[tex]\Delta KE = \Sigma W_{other}  \Delta E_{int}[/tex] Ex. Problem using this problem using the equation given above A crate of mass m = 10.7 kg is pulled up a rough incline with an initial speed of vi = 1.59 m/s. The pulling force is F = 97 N parallel to the incline, which makes an angle of θ = 19.7° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled d = 5.09 m. (a) What is the change in kinetic energy of the crate? Force done by gravity = 180J Force done by friction also the change in internal energy = 201J Force done on the crate = 494J [tex]\Delta KE = \Sigma W_{other}  \Delta E_{int}[/tex] =494J + (180J)  201J = 113J Now my question is how is that formula different from the one in my first post. Also why is internal energy considered as friction? If that is the case, what is spring potential energy? I thought that is internal energy 


#9
Dec1010, 02:46 PM

P: 1,394

So K + U = constant Ki + Ui = Kf + Uf You can see that ' ½m(v²  v₀²) + mg(h  h₀) = 0 ' is in accordance with the above equation. If you use  mg(h  h₀), it will violate the law. 


#10
Dec1010, 02:52 PM

P: 1,394

However, the workenergy theorem is still valid. So according to it, work done by all the forces equals the change in kinetic energy. Wc + Wnc + Wext = KfKi where c=conservative, nc = nonconservative, ext=external forces May be you meant work done by friction is considered as internal energy. Well if friction is an internal force in the system you consider, work done by it will count under internal energy of the system. If it is an external force, work done by it will come under external energy of the system. 


#11
Dec1010, 03:09 PM

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#12
Dec1010, 03:28 PM

P: 1,394

Using the equation Ui + Ki = Uf + Kf, Ki  Kf = Uf  Ui ½m(v₀²  v²) = Uf  Ui Now the change in potential energy of a system corresponding to a conservative internal force is defined as Uf  Ui = W, where W is the work done on the system by the internal force (here gravity is the internal force if you consider earth as a part of your system) as the system passes from its initial configuration i to final configuration f. So ½m(v₀²  v²) = [ mg(h  h₀)] ½m(v²  v₀²) + mg(h  h₀) = 0 Again in accordance with the equation I gave. 


#13
Dec1010, 03:34 PM

P: 2,568

All this internal and external system really confuses me...



#14
Dec1010, 03:57 PM

P: 1,394

A block of mass 4 kg starts falling with an initial velocity of 4m/s from a height of 10m. What is is velocity when it falls through a height of 2 m? Take the block + the earth as the system. Only the block falls, so only the work done on the block will contribute to the gravtitational potential energy. As is descends through a height 8m, the potential energy decreases by mgh. No external force is acting on the system (here gravitational force is an internal force as we already took earth in our system). So using conservation of energy, increase in kinetic energy = decrease in potential energy 0.5m(v_{f}^{2}  v_{i}^{2}) = mg(h_{i}  h_{f}) Which is same as mg10 + 0.5m(4)^{2} = mg2 + 0.5mv^{2} (taking potential energy on the surface of earth as 0) On solving, v=13.27m/s If you still don't get the concept, try reading some good books like Resnick Halliday Walker Fundamentals of Physics, Sears and Zemansky University Physics etc to develop your fundamentals. Moreover, you can't understand physics unless you think about the subject. 


#15
Dec1010, 06:45 PM

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K_{i} +U_{gi} + U_{si} + W_{nc} = K_{f} +U_{gf} + U_{sf} Where the subscripts are defined as i = initial, f = final, g = gravitational, s =spring, and where W_{nc} is the all encompassing work done by nonconservative external forces, like friction forces, applied forces, tension forces, normal forces, etc. 


#16
Dec1110, 05:26 AM

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P: 41,325

You have two choices in solving problems with conservative forces: (1) Treat them as forces: ΔKE = Work done by all forces (2) Treat them in terms of PE: ΔKE + ΔPE = Work done by nonconservative forces The first is called the "WorkEnergy" theorem; the second is called "conservation of energy". But they are equivalent (almost). In order to use these equations intelligently, you must realize, for example, that the work done by gravity is mgΔh and the change in gravitational PE is mgΔh. (The sign must change since the gravity term moves from one side of the equation to the other.) 


#17
Dec1110, 11:00 PM

P: 2,568

Shouldn't you leave it as it is and so? ½m(v²  v₀²) =  mg(h  h₀) So that ΔKE = ΔPE But it doesn't make sense when you have ΔPE  ΔKE = 0 Since I said ΔPE =  mg(h  h₀)  (  mg(h  h₀) ) ½m(v²  v₀²) = 0 mg(h  h₀)  ½m(v²  v₀²) = 0 


#18
Dec1110, 11:07 PM

P: 2,568

(2) Wait, shouldn't it be the negative of work done by all the nonconservative forces? 


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