Kdar1987 Posted November 24, 2007 Share Posted November 24, 2007 I am not great at physics... Can you help me with one problem, or just explain what should I use to solve it, which formula. Sand is dropped straight down on a moving conveyor belt at the rate of 3.0kg/s. If friction in the bearings can be ignored the power that must be expended to keep the belt moving at 2.0 m/s is: Oops.. it probably should have been in "homework help" Link to comment Share on other sites More sharing options...

JaKiri Posted November 24, 2007 Share Posted November 24, 2007 Consider what forms of energy are present in the system (hint: it's to do with what happens to the sand when it's on the conveyor). Power is the rate of change of energy, so you look at how much the energy is changing each second, and you get your answer. Link to comment Share on other sites More sharing options...

insane_alien Posted November 24, 2007 Share Posted November 24, 2007 first, work out how much energy it takes to accelerate 3kg to 2 m/s. that is the energy you must expend every second to keep it moving. E=0.5m*v^2 is the equation you should use. i'll assume you've seen it before and know what the symbols mean. Link to comment Share on other sites More sharing options...

Kdar1987 Posted November 24, 2007 Author Share Posted November 24, 2007 hmm so. for Kinetic energy I get like 6... Or should I use both Kinetical and potential? Since the answer is 12W. Link to comment Share on other sites More sharing options...

JaKiri Posted November 24, 2007 Share Posted November 24, 2007 There's no information on the potential energy, you have to assume that it's irrelevent. Link to comment Share on other sites More sharing options...

insane_alien Posted November 25, 2007 Share Posted November 25, 2007 is that ALL the information you have on the conveyor? because if it is the answer shoud definitely be 6W. Link to comment Share on other sites More sharing options...

swansont Posted November 25, 2007 Share Posted November 25, 2007 is that ALL the information you have on the conveyor? because if it is the answer shoud definitely be 6W. 12W is correct. 6W is the rate of change of the KE, but that's not the answer to the question that was asked. There's more to this than the KE of the sand. Even though the specifics aren't given (it's not necessary), friction must be present in order for the sand to move along with the belt. You have to find the force from the rate of change of momentum, and then get the rate that work is done. Link to comment Share on other sites More sharing options...

NeonBlack Posted November 25, 2007 Share Posted November 25, 2007 What level class is this homework for? In my opinion, this is definitely way too hard for highschool and probably too hard for freshman physics. As most people would, I naively expected kinetic energy [math]E=\frac{1}{2} mv^2[/math] and the power [math]P= \frac{d}{dt}E= \frac{1}{2} \frac{dm}{dt} v^2[/math] This is not quite right and here is (I think) why: (I will not be bothered with vector notation and since we are in 1 dimension anyway, it doesn't matter too much) Just using definitions: [math]F= \frac{d}{dt}p = \frac{d}{dt}mv[/math] since here, velocity is constant and mass is changing we do not have out famous result, but rather [math]F= \frac{dm}{dt} v[/math] The work: [math] W= \int{F dx}[/math] Throw in our previous expression for force: [math] W= \int{\frac{dm}{dt} v dx}[/math] Since [math]v=\frac{dx}{dt}[/math] we have [math]W=mv^2[/math] Notice that if we had the usual constant mass and changing velocity, we could integrate mv dv to get our usual result. Now using the definition of power, remembering that mass changes but velocity is constant: [math]P= \frac{d}{dt}E= \frac{dm}{dt} v^2[/math] Now the [math]E=mv^2[/math] part seems a little fishy, even to me. However, 1. I have derived it from definitions and can find no flaw in the argument. 2. (Perhaps more importantly) This gives what is supposedly the right answer. So in this democracy, we have 2 points against one. Hopefully somone else can give some input on this. Edit: It looks like Swanson posted while I was typing and confirmed that 12 Watts is correct. Link to comment Share on other sites More sharing options...

MrMongoose Posted November 25, 2007 Share Posted November 25, 2007 I don't think a physics teachers unjustified answer is more important than Newtons laws, but yeah 12W. What level class is this homework for? In my opinion, this is definitely way too hard for highschool and probably too hard for freshman physics. As most people would, I naively expected kinetic energy [math]E=\frac{1}{2} mv^2[/math] and the power [math]P= \frac{d}{dt}E= \frac{1}{2} \frac{dm}{dt} v^2[/math] This is not quite right and here is (I think) why: (I will not be bothered with vector notation and since we are in 1 dimension anyway, it doesn't matter too much) Just using definitions: [math]F= \frac{d}{dt}p = \frac{d}{dt}mv[/math] since here, velocity is constant and mass is changing we do not have out famous result, but rather [math]F= \frac{dm}{dt} v[/math] The work: [math] W= \int{F dx}[/math] Throw in our previous expression for force: [math] W= \int{\frac{dm}{dt} v dx}[/math] Since [math]v=\frac{dx}{dt}[/math] we have [math]W=mv^2[/math] Notice that if we had the usual constant mass and changing velocity, we could integrate mv dv to get our usual result. Now using the definition of power, remembering that mass changes but velocity is constant: [math]P= \frac{d}{dt}E= \frac{dm}{dt} v^2[/math] Now the [math]E=mv^2[/math] part seems a little fishy, even to me. However, 1. I have derived it from definitions and can find no flaw in the argument. 2. (Perhaps more importantly) This gives what is supposedly the right answer. So in this democracy, we have 2 points against one. Hopefully somone else can give some input on this. Edit: It looks like Swanson posted while I was typing and confirmed that 12 Watts is correct. On second thought.. If you think about it, the sand has one end at a distance v.t away from the sand input, and the input is always at 0, so the centre of mass of the sand as a whole will be half way between the two and the velocity of the centre of mass of the sand will be half the conveyor velocity. [math]F= \frac{d}{dt}p = \frac{d}{dt}mv[/math] Thats with v the conveyor velocity. Lets call the sand velocity u. [math]F= \frac{d}{dt}p = \frac{d}{dt}mu=\frac{d}{dt}\frac{mv}{2}[/math] If you folow the rest as above, the constant of a half will just fall through all the steps leading to power=6W. Link to comment Share on other sites More sharing options...

swansont Posted November 25, 2007 Share Posted November 25, 2007 On second thought.. If you think about it, the sand has one end at a distance v.t away from the sand input, and the input is always at 0, so the centre of mass of the sand as a whole will be half way between the two and the velocity of the centre of mass of the sand will be half the conveyor velocity. No, the location of the center of mass does not dictate the speed. u = v; the conveyer belt and the sand on it are ultimately moving at the same speed. 12W is the correct answer. Link to comment Share on other sites More sharing options...

MrMongoose Posted November 25, 2007 Share Posted November 25, 2007 Give me a minute. I just came up with another argument that gives me 12W. Got to work out which is wrong:confused: Link to comment Share on other sites More sharing options...

JaKiri Posted November 25, 2007 Share Posted November 25, 2007 It's a lovely question, it really gets you thinking about what is actually going on in the system. Link to comment Share on other sites More sharing options...

MrMongoose Posted November 25, 2007 Share Posted November 25, 2007 Well my previous argument holds as far as I can see other than that I forgot the second velocity should also be halved, which leads to a power of 3W. Another argument taking analogy from a jet in fluid dynamics gives me the popular 12W. The two should be the same though. Could someone please very precisely tell me why my first argument was wrong and revive my faith in fluids? Link to comment Share on other sites More sharing options...

swansont Posted November 25, 2007 Share Posted November 25, 2007 Could someone please very precisely tell me why my first argument was wrong and revive my faith in fluids? Do you mean the one where you claimed that the sand is moving at half the speed of the conveyer belt? I addressed that above. They move at the same speed. Link to comment Share on other sites More sharing options...

MrMongoose Posted November 26, 2007 Share Posted November 26, 2007 I do mean that, but I don't feel you addressed the bit thats getting me. I DO understand that all the sand that was on the conveyor before the point in time we look at it is irrelevant as its momentum is constant and that if we look at elements of sand as theyre added we get the right answer, and I don't dispute that. At the same time, I don't see why a line of sand cant be modelled as a particle at its centre of mass (which moves at v/2) with mass equal to the total mass of sand. Is that not similar to modelling a rod that undergoes no moments as a particle in a dynamics problem such as a rod falling through the air under the air at terminal velocity? Link to comment Share on other sites More sharing options...

swansont Posted November 26, 2007 Share Posted November 26, 2007 I do mean that, but I don't feel you addressed the bit thats getting me. I DO understand that all the sand that was on the conveyor before the point in time we look at it is irrelevant as its momentum is constant and that if we look at elements of sand as theyre added we get the right answer, and I don't dispute that. At the same time, I don't see why a line of sand cant be modelled as a particle at its centre of mass (which moves at v/2) with mass equal to the total mass of sand. Is that not similar to modelling a rod that undergoes no moments as a particle in a dynamics problem such as a rod falling through the air under the air at terminal velocity? The question is asking how much power it takes to accelerate the sand from rest to speed v. The average speed of the particles still undergoing acceleration isn't pertinent. You are looking at the behavior of an arbitrarily small mass element, dm. The line of sand can be arbitrarily long, and the bulk of it moves at v. It is simply incorrect to claim the COM moves at v/2. Sorry of that sounds brusque. I can't identify the misconception in play here. Link to comment Share on other sites More sharing options...

MrMongoose Posted November 27, 2007 Share Posted November 27, 2007 I think basically I've been hanging onto a simplification my sixth form maths teacher made for too long, and a lot of ponderign has set me straight! Thanks for letting me think out loud! Link to comment Share on other sites More sharing options...

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