wel its abt a thing) i guess i\nhave found. i think there is a way by which u\ncan disprove (or violate ) the law of conservation(L.C.E)\nof energy\nby showing potential energy not = kinetic energy. its a\ntricKy situation where only law of conservation of momentum is\nright OR L.C.E WORKS.wel iv been thinkin abt it for the past 4\nyears ..i tried tellin this my friends nd they didnot\nlisten..so i decided 2 start a group.\nNOW ILL TEL U HOW.\ntake a long steel chain (made of oval shaped\nlinks...i.e.. a cylindrical steel rod bent into the form of an oval shaped link.and 1 metre of that chain should weigh 1 k.g )of length\n85metres(the longer the better) . take the chain to a platform\nof height of say 5 metres. now arrange the chain in such a\nway that when u drop one end of the chain down 5\nmetres ,rest of the chain flows down freely like a stream of\nwater.(i mean that the chain shouldnot b entangled  )\nnow drop one end of the chain so that it\nhits the ground.this whole process is as frictionless as\npossible.the chain will start to flow like a stream of water\nand it will take a few seconds for the other end to hit\nthe ground.\nnow how long will it take for the entire length of chain\nto run down like a thread ??. as per calculations ,(taking\nacceleration due to gravity as 10 metres per sec per sec .and there\nis zero air resistance .the process is considered as\nfrictionless as possible.) the time it will take for 80 metres\nof chain to run down and hit the ground will be 11.315\nseconds . and at the piont of hitting the ground every part of\nthe chain will have a velocity of 7.07 metres per second. so\ntotal K.E OBTAINED when 80 metres of chain rundown will be\nK.E = 1/2 *M*V*V\n= 1/2 *80*7.07*7.07\n= 2000 UNITS .\nWHICH IS half the energy required to lift 80 metres of chain\nto a height of 5 metres as given below.here 7.07metres per\nsec is the terminal velocity when height is 5 metres.{when\nheight is 20m term velocity is 14.14 m per sec}.\nhere law of conservation of momentum works but not L.o.C.E..\n\n((when u drop all of the chain together it will hit\nthe ground in 1 seconds ( take acceleration due to gravity as\n10 m per sec per sec). at time of hiting the ground every\npart of chain will have a velocity of 10m per second.this way\nP.E = K.E .(P.E = M*G*H, K.E 1/2 *M*V*V)\nI.E., P.E = K.E.\n80*10*5 = 1/2*80*10*10.\n4000 units = 4000 units.))\n\nTHIS EXP  CAN BE   DONE    USING   MUCH LONGER  LENGTHS OF  CHAINS.\nTHE  HEIGHT FROM  WHICH U  DROP  IT CAN  ALSO  BE  VARIED ..\nmy contact address is  mr balaji.k no 5, balaji illam , postal colony ,p.n road ,tirupur-641602 india. ph no- 00 91 98430 53634
130726
You seem to assume that every link in the chain will have the same velocity when it hits the ground.  However, that will not happen.  Each link is pulled by the previous link such that it begins its descent with an initial velocity higher than the one before it.  Then, it is accelerated due to gravity, thereby increasing the initial velocity transfered to the next link.  In case this is not clear, let me explain further:  Note that each link in the falling part of the chain has to be going at the same velocity since they are connected to each other.  If they were moving at different velocities, they would quickly separate each other from themselves.  Further note that the speed they are falling has to be increasing because of gravity.  Therefore, links higher up in the chain do not start falling from rest; they begin falling with an ever increasing initial velocity.\n\nThe experiment you propose is equivalent to holding the chain straight up and dropping it from a certain height (you used 5 meters).  The initial velocity that each link imparts on the next link in the chain is the same as it would be had the link fallen on its own the entire time.  This is because  in your experiment, each link is given an initial velocity that is equivalent to that link's having been accelerated due to falling for the entire time that the chain had been dropped.  \n\nThe only difference in the experiments is the form of energy the links have.  In your experiment, the links begin their descent with some kinetic energy and some potential energy and end their descent with purely kinetic energy.  In the equivalent experiment, the links begin their descent with purely potential energy and conclude it with purely kinetic energy.  In either case, a link in one chain will have the same amount of energy as the same link in the other chain because the two experiments are equivalent.\n\nAssuming you agree that the law of conservation of energy holds for a chain being held straight up, you should be able to accept that it also holds in the experiment you proposed.
340816
Let v1 = speed of the plug relative to a dot\nLet v2 = speed of the infinite pipe relative to a dot\nLet phi = angle of the pipe relative to some plane\nspeed of the pipe along the plane is v2 * Cos(phi)\nphi = 0 if we select a plane along the axis of the pipe\nTherefore the vector describing the speed of the pipe along the plane is v2*Cos(0) = v2\nThe plug speed relative to the plane is also found to be v1\nAn infinite pipe will act like a line that is very far away\nThe plug will act like a dot moving along that line if it is very far away too\nAssume that our origin is very far away from the pipe - if the pipe is infinite then space has to be very big and the assumption is correct for all 3-dimensional space.\nThe speed of the plug relative to the pipe is v1-v2\nv2 = 0 since if it was greater than 0 then it would have taken an infinite amount of energy to move it since it would have an infinite mass and no energy would have been left over to make the plug.\nSince the ends of the pipe are infinitely far away then the pipe must have come into existence after the plug\nThe guy who got the pipe wrapped around the plug would never have been able to do it if the plug was moving\nTherefore v1 = 0\nIf v1=0 and v2=0 then v1=v2\nSo in fact the plug is stopped relative to the pipe.
101638
It's at the apex of the circle.\n\nThe total energy of the system is the sum of the kinetic energy, 1/2 mv^2, and the potential energy, mgh (where h is the height above the circle's lowest point).  At the bottom of the circle, all the energy is kinetic and the speed is highest; at the top, some of this energy has been stored as potential energy, according to this:\n\n1/2 m (v_bottom)^2 + 2mgr = 1/2 m (v_top)^2 \n\nThe actual value of the velocity is undetermined, as v_bottom can be as fast as you like.\n\n>>>EDIT:\n\nPoster above is completely right.  Whoops :)  That'll teach me to do this on scraps of paper at work...
374788
An object with momentum always has energy.  Not all objects with energy have momentum.\n\np = m*v \nWhere:\np - momentum\nm - mass\nv - velocity\n\nIf velocity (v) is 0, then p = 0\n\nConversely, any object that is not at rest has momentum.\n\nRegarding energy, E = m*(c^2)\n\nTherefore any object of mass has energy even at rest.  But still without velocity there can be no momentum.\n\nOn a side note, for all you relavist out there, please remember the defining principle of momentum...it is something which must be able to be transferred within a single relative frame of reference.  Therefore, when we talk about velocity, we must define the point of reference.  A coin sitting on a table has no momentum because in reference to the table, it has no velocity.  This statement is true even though the table rests on the floor which rest on the foundation which is embedded into the soil which is part of the earth's crust which is revolving on the earth's axis at 24000 miles per hour relative to the movement of the sun.  How you like them apples?
474473
We place the weight of the rod at the center of its length or 0.90m from one end. Its Potential Energy, PE at the horizontal position with respect to the particle, is 2.4g*1.40.  This energy is converted to a lesser PE at theta=35 degrees.  PE of the rod at this point is          2.4g*(1.40-0.90cos35). The particle had initial PE equal to zero, but at 35 degrees from its original position, it gained a PE equal to mg(1.40-1.40cos35). The PE gained by the particle is equal to the dissipated energy  during the collision. \n\nBased on the principle of conservation of energy:\n\nPE of system, initial=PE of system, final\n\na)Substitute the above values:\n\n2.4g*1.4=2.4g(1.40-0.90cos35)\n+mg(1.40-1.40cos35)\n\ng cancels out.\n\n2.4*1.4=2.4*0.67+0.26m\nm=(3.36-1.60)/0.26\n   =6.76kg\n\nb)Dissipated energy=6.76g*0.26 \n                           =17.2 Joules\n\nPls draw a diagram showing the position of the rod and the particle, before and after collision. You'll then find my explanation easier to understand.
468744
E, G, J\n\nin this problem, it can be assumed that time of fall is of sec's only hence air resistance can be negligible. problem can be solved using hooke's law though i am not sure what principle bungee rope works. heant generation which will take place in rope will be much smaller then actual work done. There can't be a constant speed or accelaration coz both r changing constantly, for example to start of jump speed is zero, acc downwards, where as at bottom point speed is again zero but acc. is upwards. n in between speed is non zero. \nlast, u can treat him as a particle for simlification as long as u keep air resistance negligible. u must realize that neeta (coff. of air resistance is very small n play effective role at high speeds).\ncord applies different force at different moments becoz of hooke's law.\n\nI hope it helps......
92437
You have some very good answers here but I can add a little more.  Consider a pipe to be elastic (true!) but imagine that it is made of rubber.  Consider water to be nearly incompressible (a cubic inch of water at the bottom of the 7 mile deep Mariana trench still measures nearly a cubic inch).  Water can possess energy in two ways (neglecting heat); pressure energy or velocity energy and can easily convert back and forth.  When you extinguish velocity of a slug (or stream) of water you produce pressure and the pressure when relaxed can produce velocity.  This can go on and on until the energy is dissipated for example by friction.\n\nWhen a stream of water is flowing along an elastic pipe (think rubber) and you suddenly trip shut a guillotine valve (very fast acting) the velocity of the column of water is instantly extinguished and converted to pressure which starts as a shock wave stretching the rubber pipe toward the upstream end perhaps until it reaches the vessel supplying the water.  But the stream is then reversed as the rubber pipe relaxes squeezing out the water toward the vessel and continues until the rubber pipe is under a partial vacuum and shrinks.  In an ideal system (no friction), pressure waves would continue the travel back and forth along the pipe converting pressure to velocity and back to presure as the pipe expands and contracts under the changing pressure.  They call it water hammer because it actually can sound like someone hitting the pipe with a hammer.  Actually the pipe expands suddenly sending off a sound wave through the air.\n\nWater hammer can be avoided several ways.  For example if the pipe also contains air bubbles (accidently or on purpose) the bubbles will absorb any hydraulic shock wave preventing the elastic pipe from expanding and the bubbles re-expand slowly, slowing down the entire procedure.  The plumbing leading to many sinks in homes has a small "empty" vertical section near the sink to hold an air bubble which avoids a water hammer if the sink faucet is shut too fast.  It is sometimes necessary to drain the system with valves open to refill the air trap with air if it gets waterlogged.\n\nWater hammer can be avoided by shutting valves slowly rather than instantly.  Water hammer can occur in a pipe filled with steam because steam is easily compressed compared to an air bubble and a slug of water can be accelerated along a pipe by condensing steam until it "strikes" the end of the run and converts velocity to over-pressure.  I once got a frantic call at home (3 AM?) from a power plant under construction.  While testing a large pipe run for the first time under steam, the long say 12-inch pipe began to jump around threatening to break its pipe hangers and making a terrific racket.  Almost without waking up I told them to crack open any lowpoint valves or fittings to drain off all condensation and the pipe went back to sleep too.  In the finished plant there were automatic condensation drain valves to avoid water hammer which could damage the steam turbines too, etc.
11451
nothing's wrong with the above and your conversion is correct.\n\ndoesn't seem like very much eh?  the reason is b/c of what i explained in the other question.. if you plot a graph of the force versus time, it will look like this:\n\n_____|___|___|___|___|___|___|___|_____\n\nthose sharp vertical lines represent the bullet impacts, and the rest of the time the force is zero.  if you average over this time span, the number turns out to be very low.  but the height of those peaks is very high because so much momentum is exchanged in a very short time!  so force is kind of misleading.  in physics the more meaningful way to talk about collisions is to talk about the momentum exchanged in a collision, which is called an "impulse" because it happens over such a short time.\n\nput another way, the bullet has a momentum of m * v, or .003 kg * 740 m/s = 2.22 kg*m/s.  think about what that means.  it's like getting whacked in the chest with a 2.22 kilogram (around 4.5 lbs) object moving at 1 meter per second - pretty heavy and pretty fast!  \n\nand that's just one bullet.  think about 8 of them hitting you every second.  furthermore, 3 grams seems like pretty low estimate for the mass of a bullet.   that's something like a .22 or a .25.  consider that a .44 magnum bullet, which is the sort of gun that actually can knock you off your feet, has a mass of 250 "grains" which is closer to 16 grams.  that's more like getting hit with a 10-kg object.  that's like someone winding up and throwing a 22-lb weight at you!\n\nso there you go.  the averaging and the short time interval for the impulse is what's misleading you.
387755
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
528811
You're in luck.  I teach physics and I know exactly how to answer this question (it's based on the work-energy principle).  It's pretty simple, really.  But it would be best if we discussed it by messenger.  Yahoo Messenger me, my ID is\nfortitudinousskeptic\n\n- Kevin
19029
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
340822
In conclusion we found the water to bead up like a dome until the increase in mass/weight overcame the Penny's ability to hold the water. Then the water ran onto the table across the floor. There was a bare electrical wire which the water shorted out and it caught fire, burning the building down. After the Fire Department left, we decide never to try this stupid experiment again!\nYou may want to reword it some, this this best I could do from what you wrote.
12544
what your calling energy is really Net Force.  or Mass x Acceleration.  \n\nin the first example, the energy wasn't gone, if you assume that the people have the same mass and were traveling at teh same velocities when they reached teh end then the net force acting on the system is 0 because the force (ma) of the first person is equeal to the force of the second person, so they cancel.  \n\nin the second example, a centrepedial force, or force towards teh center, is being applied from teh rope on the people holding it.  this causes the direction of their velocity to change.  using dot products we can find teh work done, which is change in kinetic energy.  you will find that the work done on one person is equal and opposite to the work done on the other.
322855
Kinetic energy Ek\n\nlet's consider that an object with a mass m moves on a distance d = x1 - x2, with an acceleration a and its velocity changes from v1 to v1, under the action of a force F\n\naccording to Galileo's law:\nv2^2 - v1^2 = 2.a.d\n\nbut according to Newton's second law:\nF = m.a\n\n-> a = F/m \n-> v2^2 - v1^2 = 2.F.d/m\n-> m.v2^2 - m.v1^2 = 2.F.d\n-> m.v2^2/2 - m.v1^2/2 = F.d\n\nbut W = F.d = the work done while the force F moves the object on the distance d, it caracterizes a process\n\nBut energy is the capacity of an object of doing work, it caracterizes a state\n\nthe kinetic energy is due to the velocity of the object:\n\ndeltaEk = E2 - E1 = W = F.d \n-> E2 = m.v2^2/2\n-> E1 = m.v1^2/2\n\nPotential energy Ep\n\nLet's consider an object with a mass m is falling from a height h1 to h2. \n\nThe work done by gravity is:\n\nW = G.(h1-h2) = m.g.(h1-h2) = m.g.h1 - m.g.h2\n\nThe potential energy is depending on the position of the the object as compared to the Earth and its equal with the work done by gravity taken with opposite sign.\n\ndelta Ep = Ep2 - Ep1 = - W\n\n-> Ep2 = m.g.h2\n-> Ep1 = m.g.h1
333972
I don't usually answer homework questions, but you're being led astray by kingkongchess, and you'll pay the price at test time anyways if you don't figure these out for yourself, so I'll answer one and give you some hints on where to look for with the others.\n\n1) F=ma=dp/dt=m x dv/dt = constant, so m x delta v=const. x delta t.  More simply, m x v=const. x time, but be warned, this WONT work unless the force is constant!  Anyways, v=(32N x .16s)/.127kg = 40.3 m/s (just divide away the mass for the velocity).  A little more likely than 1575 m/s (~3500 mph!), no?  That'd be some hockey player!\n\n2) kingkongchess sets it up exactly right; all answers come from the conservation on momentum, m1v1=m2v2.  This is also true for problem 3.\n\n3) Trickier.  Before collision, the system consists of 2 seperate momentums, p1 and p2.  After the collision, the system consists of the single body (with momentum (m1 + m2) x v3 ) stuck together.  Conservation of momentum states that the momentum of the system before the collision must equal the momentum of the system after collision.  Just make them equal, and solve for what you're looking for.\n\n4) The word 'accelerates' should give you a clue.  Think 'force' when you hear the word accelerates, and when you think force, think 'rate of change in momentum'.  Put these ideas together, you'll have your answer.
540877
This would potentially be a very complex problem if the blocks did not have the same mass and/or the same initial speed. The reason is that, since there is no external force on the system, its center of gravity remains stationary - but if the blocks began to move at different speeds, it would be difficult to keep track of that.\n\nAs it is, you can argue by symmetry and the above observation that the blocks will decelerate equally and always remain symmetric about the center of the spring.\n\nTherefore all the kinetic energy of the blocks is converted to potential energy of the spring: 1/2 (2 *m) v^2 = 1/2 k x^2\n\nSince you know all the numbers it is easy to work out that the maximum compression is .0156 m or 15.6 mm.
91871
see if this helps.....\n\n\n\nAnalysis of a simple gravity pendulum\n \nFigure 2. Force diagram of a simple gravity pendulum.To begin, we shall make three assumptions about the simple pendulum\n\nThe rod/string/cable on which the bob is swinging is massless and always remains taut; \nThe bob is a point mass; \nMotion occurs in a 2 dimensional plane, i.e. pendulum does not swing in and out of the page. \nConsider Figure 2. The blue arrow is the gravitational force acting on the bob, violet arrows are that same force resolved into components parallel and perpendicular to the bob's instantaneous motion, the motion along the red axis, which is always perpendicular to the cable/rod. Newton's second law\n\n \nwhere F is the force acting on mass m, causing it to accelerate at a meters per second2. Because the bob is constrained to move on the green circular arc, there is no need to consider any force other than the one responsible for instantaneous acceleration parallel to the known path, the short violet arrow in our case\n\n \n \nLinear acceleration a along the red axis can be related to the change in angle θ by the arc length formula\n\n \n \n \nThis, however, is not the acceleration we seek because the gravitational force on the bob causes a decrease in angle θ. One accounts for that by placing a negative sign in front of a, thus:\n\n \n \n \nFigure 3. Trigonometry of a simple gravity pendulum.This is the differential equation which, when solved for θ(t), will yield the motion of the pendulum. It can also be obtained via the conservation of mechanical energy principle: any given object which fell a vertical distance h starting from rest, would have acquired kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy. Change in potential energy is given by\n\n \nchange in kinetic energy (body started from rest) is given by\n\n \nSince no energy is lost, those two must be equal\n\n \n \nUsing the arc length formula above, this equation can be re-written in favor of \n\n \nbut what is h? It is the vertical distance the pendulum fell. Consider Figure 3. If the pendulum starts its swing from some initial angle θ0, then y0, the vertical distance from the screw, is given by\n\n \nsimilarly, for y1, we have\n\n \nthen h is the difference of the two\n\n \nsubstituting this into the equation for  gives\n\n \nThis equation is known as the first integral of motion, it gives the velocity in terms of the location and includes an integration constant related to the initial displacement (θ0). We can differentiate, by applying the chain rule, with respect to time to get the acceleration\n\n \n \n \nsame as obtained through force analysis.\n\n[edit]\nSmall angle approximation\nThe problem with the equations developed in the previous section is that they are unintegrable. To shed some light on the behavior of the pendulum we shall make another approximation. Namely, we restrict the motion of the pendulum to a relatively small amplitude, that is, relatively small θ. How small? Small enough that the following approximation is true within some desirable tolerance\n\n \nif and only if\n\n \nSubstituting this approximation into (1) yields\n\n \nSolution to this equation is a well-known, and quite expected, oscillatory function\n\n \nwhere θ0 is the semi-amplitude of the oscillation, that is the maximum angle between the rod of the pendulum and the vertical.\n\nThe term  is a pulsation, which is equal to ,\n\nwhere T0 is the period of a complete oscillation (outward and return).\n\nSince ,\n\nthe period of a complete oscillation can be easily found, and we have obtained Huygens's law:\n\n \n \n[edit]\nArbitrary-amplitude period\nFor amplitudes beyond the small angle approximation, one can compute the exact period by inverting equation (2)\n\n \nand integrating over one complete cycle,\n\n \nor twice the half-cycle\n\n \nor 4 times the quarter-cycle\n\n \nwhich