albertlee Posted February 16, 2005 Posted February 16, 2005 Why we cant express Work interms of Momentum??? One is F*d and another F*t..... Albert
swansont Posted February 16, 2005 Posted February 16, 2005 You can. Work is change in kinetic energy (if there is no potental energy change), and KE = p2/2m
albertlee Posted February 16, 2005 Author Posted February 16, 2005 Why work = F*d??? I dont see there is no reason why we cant express work such that it equals to F*t??? Is there a relationship between work and change in momentum?? Albert
mezarashi Posted February 17, 2005 Posted February 17, 2005 Well, that's a very unusual question you have. Work and Momentum are concepts used in physics that can indeed help us to model and predict our world. Their notions to exist as with any other model in physics is because it is a tool for us in problem analysis. Work is the concept of looking at the world in terms of relative energy content and energy exchange (i.e. the concept of conservation of energy) Momenum works on its own concept of conservation of momentum, or more like the conservation of inertia. Your question, Why is work F*d and not F*t... well if they were both the same, we wouldn't have to come up with two words and two concepts for them ^^;. Although the two models are usually not used together in most cases. I don't think there is a general relationship, as it depends on the situation. In simple linear motion however, I think it may be safe to say that generally a loss of momentum also constitutes a loss of kinetic energy. Feel free to ask anything else
swansont Posted February 17, 2005 Posted February 17, 2005 Why work = F*d??? I dont see there is no reason why we cant express work such that it equals to F*t??? Is there a relationship between work and change in momentum?? Albert We can't because work is defined as it is' date=' and momentum is defined as it is. I already gave an expression for KE and momentum. You can't easily relate it to a change in momentum because KE depends on p[sup']2[/sup], so it depends on how much momntum you have to start with.
albertlee Posted February 17, 2005 Author Posted February 17, 2005 This is just how I think: Force * Distance Force * Time If you have a force exerting on an object going for a longer distance, more work will be done..... Similarly, if a force exerting on an object going for a longer period, more work will be done as well.... To me, they both "can" express work done. It is simply saying like in terms of time or in terms of distance...... To me, impulse, the change in momentum of an object actually tells how much work has been done, ie, how much does the object accelerate, and of course, the longer the object accelerates, the longer its distance will be, right?? Albert
Primarygun Posted February 18, 2005 Posted February 18, 2005 In fact, momentum is usually used to describe the movement of colliding objects. If time increases, displacement increases(take absolute value).If displacement increases, more time is needed, holding other variables being constant. In general, if you can't describe work done by momentum in some difficult or easy ways, then it is not mechanics. They are always related as you have mentioned. That's my thought.
albertlee Posted February 18, 2005 Author Posted February 18, 2005 Well, What I really want to know, is, since F*d is used as work done, there must be more ideas/concepts apart from momentum interms of work done..... It is hard to give a plain explanation to momentum other than mathematical equation, but to me, it is the "quantity of motion" In one sentence, why not the scientist let "quantity of motion" = work done???? Albert
Primarygun Posted February 18, 2005 Posted February 18, 2005 So you mean use Fd to describe the motion of an objects? What if an object moves with constant velocity? There is no "external net force" acting on it. Fd is better used to describe the "potential" of an object.
swansont Posted February 18, 2005 Posted February 18, 2005 This is just how I think: Force * Distance Force * Time If you have a force exerting on an object going for a longer distance' date=' more work will be done..... Similarly, if a force exerting on an object going for a longer period, more work will be done as well.... [/quote'] But how long you exert the force does not measure how much energy you have given it. As I said before, you can figure out the momentum from the kinetic energy, and vice-versa. But KE and momentum are not the same thing.
Primarygun Posted February 18, 2005 Posted February 18, 2005 I think he thinks that creating a new term " momentum " is not necessary.
fermions Posted February 18, 2005 Posted February 18, 2005 I think that KE and momentum is considered as one whole thing when considering four dimensions... but I'm not sure about it...
timo Posted February 18, 2005 Posted February 18, 2005 it´s not exactly the kinetic energy only but the total energy. The vector containing energy and momentum is then called the 4-momentum.
albertlee Posted February 19, 2005 Author Posted February 19, 2005 Can any one show me the step to derive to K.E. = mv2/2 from F*d?? thx
The Rebel Posted February 19, 2005 Posted February 19, 2005 Can any one show me the step to derive to K.E. = mv2/2 from F*d?? thx Imagine a car that is being accelerated from rest. The engine is doing work which is directly providing an increase in KE. So' date=' Change in KE = W = F*d = mad One of the velocity formulas that relates to changes in velocity is: v[sup']2[/sup] = u2 + 2ad, which reduces to v2 = 2ad as u = zero. therefore Change in KE = mv2 / 2
Tom Mattson Posted February 20, 2005 Posted February 20, 2005 Can any one show me the step to derive to K.E. = mv2/2 from F*d?? Sure' date=' why not? I need to practice my LaTeX. The following calculation is the case of a particle moving in 1D. [math'] W=\int_{x_1}^{x_2}Fdx [/math] [math] W=m\int_{x_1}^{x_2}\frac{dv}{dt}dx [/math] [math] W=m\int_{x_1}^{x_2}\frac{dv}{dx}\frac{dx}{dt}dx [/math] [math] W=m\int_{x_1}^{x_2}\frac{dv}{dx}vdx [/math] [math] W=m\int_{v_1}^{v_2}vdv [/math] Note that in the last step I changed the limits of integration, which is because I changed the variable of integration. Finally: [math] W=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 [/math] If you start from rest then v1=0 and you can simply set v2 equal to v, since there will no longer be any need to distinguish the two. Then you have the formula for KE. BTW: Cool avatar, albert.
albertlee Posted February 20, 2005 Author Posted February 20, 2005 Well, I want to know the derivation of K.E. not change in K.E. Any way, how come I got different result for v2??? we all know that v = u + at t = d/[(v-u)/2] = 2d/(v-u) v = u + a[2d/(v-u)] = u + 2ad/(v-u) v2 - uv = vu - u2 + 2ad v2 = 2vu - u2 + 2ad Any help to the manipulation??? Secondly, I find it very helpful to see the relationship between momentum, force, and work in this way: momentum = kgm/s = Ns force = kgm/s2 = N work = kgm2/s2or kgm/s2(m) = Nm Ps. Tom, I cant understand your derivation, because I haven't learned calculus, or differential calculus or what ever it is, yet, Albert
swansont Posted February 20, 2005 Posted February 20, 2005 Well, I want to know the derivation of K.E. not change in K.E. That is the derivation. Start with v=0. t = d/[(v-u)/2] = 2d/(v-u) This is undefined if v=u. So an object travelling at constant speed takes an infinite time to travel a distance d. Is that reasonable?
Tom Mattson Posted February 20, 2005 Posted February 20, 2005 Well' date=' I want to know the derivation of K.E. not change in K.E.[/quote'] As I said, and as swansont pointed out, you get the formula for KE if you start from v1=0. Any way, how come I got different result for v2??? we all know that v = u + at t = d/[(v-u)/2] = 2d/(v-u) First of all, what you are trying to do is only valid for motion with constant acceleration. That won't get you from the definition of work to the formula for KE. Second, in the case of constant acceleration you should divide t by the average speed, which is: [math] v_{avg}=\frac{v+u}{2} [/math] as opposed to what you have in your last line. Secondly, I find it very helpful to see the relationship between momentum, force, and work in this way: momentum = kgm/s = Ns force = kgm/s2 = N work = kgm2/s2or kgm/s2(m) = Nm Well, that's good. You should always check your units. But units alone will not help you understand why work is defined the way it is, or why energy and momentum are both conserved quantities. Ps. Tom, I cant understand your derivation, because I haven't learned calculus, or differential calculus or what ever it is, yet, Sorry, the definition of work is in terms of an integral. Can't do much about that.
albertlee Posted February 20, 2005 Author Posted February 20, 2005 Oh... made a mistake that average velocity = v+u/2 not v-u/2 The law of conservation of energy states that energy can never be destroyed, it only transforms into different forms, how about the law of conservation of momentum?? thx Albert
timo Posted February 20, 2005 Posted February 20, 2005 Conservation of momentum is the same; total momentum is conserved. In fact -as it was allready noted- energy and momentum can be seen as very similar. In relativity they are both elements of the 4-momentum which is conserved. The 4-momentum is p = (E, px, py, pz).
albertlee Posted February 20, 2005 Author Posted February 20, 2005 Ok, can you explain the conservation of momentum plainer?? By the way, what is the similarity of conservation of momentum and conservation of energy?? as Atheist noted?? Albert
Primarygun Posted February 21, 2005 Posted February 21, 2005 Both must be conserved. Law of conservation is totally supported by Newton's third law.
albertlee Posted February 21, 2005 Author Posted February 21, 2005 What does that explain??? When A exerts a force on B, B exerts an equaled magnitude force but in opposite direction What does explain the law of conservation of momentum??? Albert
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