If its true that energy isn't lost, its just transferred, where did the energy go from a falling object that hits the floor and stays there? It started with the most gravitational potential energy, a second before it lands it is converted to kinetic energy, and the kinetic is at the peak. But after it fell and stopped, where is all the energy?  Also, it is said you can't create energy. But when an object is at rest on the floor it has no energy, and after being picked up it has energy.  I know these are basic questions, but I can't get to the bottom of this.
123776
Kinetic energy conservation in collision.  In any type of collision type of collision if we take the system to comprised of both bodies then the net external force is zero. So the work done by external force is also zero. By the relation $W=K_f-K_i$ as $W$ is zero hence kinetic energy is conserved. But this not the case.   Can you tell me what is the flaw in my argument? Some people say that the argument has neglected the fact that some kinetic energy is transformed into heat. But heat is created due to the interactions between bodies during collision which is thus an internal force and thus shouldn't affect kinetic energy.
18737
Take a glass of water and piece of toilet paper. If you keep the paper vertical, and touch the surface of the water with the tip of the paper, you can see the water being absorbed and climbing up the paper. This climbing up is intriguing me. Where is the energy coming from?  Supposedly, a lot of water molecules are moving up and gaining potential energy. This has to be balanced by something. I can think of a few possibilities, but I can't tell which one it is.   When the water molecules dilute into the paper, they are at a state of lower potential binding energy. Some molecular interaction (van der Waals?) is at lower energy when you have a water+paper solution, compared to a water-only solution. This compensates the gain in gravitational energy. The surface between the paper and the water is at lower pressure than the atmosphere. This causes the water to be pushed into the paper by the atmospheric pressure, up until the point that the column of water above the surface is heavy enough to counterbalance. The potential energy would be a result of work done by the atmosphere. Some water molecules climb up randomly and loose kinetic energy going up, and somehow get "stuck" up there. Something else?
83860
I am stuck in a question where energy conservation is failing and momentum conservation is correct. I think I might be doing something wrong, that's why I'm asking this question. The problem is as follows:     A bullet of mass $25\,g$ is fired horizontally into a ballistic pendulum of mass $5.0\,kg$ and gets embedded in it. If the centre of pendulum rises by a distance of $10\,cm$, find the speed of the bullet.   (H.C. Verma, Centre of Mass, Q47)     Let the mass of the bullet be $m=25\,g$. Let the mass of the pendulum be $M=5\,kg$. Let the peak height be $h=0.1\,m$. Let the initial and final velocities be $u$ and $v$ respectively.  Method 1: (Momentum Conservation)  $$mu=(M+m)v \implies v=\frac{mu}{M+m}\\ \text{Also, }\frac{1}{2}(M+m)v^2=(M+m)gh\\ \implies u^2=2\biggl(\frac{M+m}{m}\biggr)^2gh\\ \implies u=201\sqrt2 \text{ m/s}$$  Method 2: (Energy Conservation)  Since both the masses move together after the collision, and because the velocity at highest point is null, therefore: $$\frac{1}{2}mu^2=\frac{1}{2}(M+m)v^2=(M+m)gh\\ \implies u^2=2gh\biggl(\frac{M+m}{m}\biggr) \implies u=\sqrt{402}$$    To summarize myself, I am curious about the following: Why do the results differ? Shouldn't the energy be conserved as well as the momentum? Since it should be, then what is the flaw in my calculations?
115118
If using the surface of the earth as a reference point how much work is needed for gravity to pull me to the center. Is it negative infinity or am I wrong? Also is a single value of potential meaningless. Does it represent anything? Do you at least need a potential difference for it to represent work done? I'm extremely confused about potential and potential energy.
66967
An everyday is example would be cars. If you were to drop a car, not taking air resistance into account, would a front-heavy car tumble forward as it fell? And a car with a perfect 50/50 front/rear weight distribution fall straight? Does mass distribution/center of mass affect free fall? Or would they both fall straight down without tumbling? Can you explain why please?
6494
I know this question is asked here a lot, but I just had to ask this to finalise the concept.  When a system lets say a rod of length $L$ and mass $M$ is rotating with angular speed $omega_1$ its initial angular momentum is $L1 = (1/12)ML^2\omega_1$ and its initial kinetic energy is $KE = (1/24)ML^2{\omega_1}^2$.  Now after some time the rod is folded in half its angular momentum kept conserved i.e. without applying any external force or torque, its new angular velocity becomes $\omega_2 = 4\omega_1$ and its new kinetic energy becomes $KE_2 = (1/6)ML^2{\omega_1}^2$.  This is 4 times the original kinetic energy when no external force works, since the rod is folded, you can even say melted and formed into a smaller and denser rod, it has not undergone compression/expansion of any sort but still there is change in kinetic energy.   The most sense I could make out of this was that all the particles while rotating felt a centripetal force and the particles of half of the rod under this force went in its direction and did some work which appears as the change in kinetic energy. I have written the same and a proof as an answer .  Now if I am write in my concept where did this energy come from, tension was providing the centripetal force but no work was done against tension as the rod was folded in half not compressed. If I am wrong then where did the energy come from?  Extra: I also tried the analogy of this question in translatory motion, suppose there is body of mass $m$ moving with velocity $v$  suddenly, its mass becomes $m/2$ then its velocity becomes $2v$ and its KE becomes $4KE_1$ there is no need to explain the energy conservation here since mass suddenly does not disappear into thin air, however moment of inertia can be changed and hence the question.  Addendum : Since folding the rod seems to bring about unnecessary questions about ways of folding, you can imagine that if rod was melted and formed into a longer rod all the while the system was rotating and angular momentum conserved, then new length becomes $2L$ new angular velocity becomes ${\omega_1}/4$ and new KE becomes $(1/96)ML^2{\omega_1}^2$. This time the energy becomes ${1/4}^{th}$ of the initial, where did this energy go to ? Certainly movement against centripetal force takes place, but since there is no extension in existing rod, energy can not be stored as spring energy in it, or so I think.
45936
Why is there conservation of kinetic energy in  and not in ? What would be the difference that lead to conservation of kinetic energy in elastic collision and not in inelastic collision? We also know that mechanical energy is not converted to heat, light, sound etc. in elastic collision but converted to heat, light, sound etc. in inelastic collision. Why is mechanical energy converted as total energy is conserved in inelastic collision?
112778
Let's start with kinetic energy (from )     The kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed $v$ is $\frac{1}{2}mv^2$.   Let's say you &amp; your bike have mass of 100kgs, then your kinetic energy at 10m/s would be  $$ E_a = 1/2 \times 100 \times 10^2 = 5000J = 5kJ$$  If you apply another 5kJ of energy, you don't get to 20m/s though, you only get to:  $$ E_b = 10000J =1/2 \times 100 \times V_b^2$$    $$\implies V_b = \sqrt{10000 / (1/2 \times 100)}         = √200         = 14.14m/s$$  Let's say you and a buddy are both coasting along at 10m/s though, from their perspective you've just burned 5kJ but only accelerated 4.1m/s, even though you seemed stationary.  Imagine you and your mate are in space drifting along together, at an unknown speed. Your mate fires his burners and accelerates away from you. There's a big screen on his ship showing how many joules of energy he just burned, and you can measure his resulting relative velocity just fine.  The question is, Will 5kJ of energy always produce 10m/s of relative velocity,,assuming 100kg spaceships?  If 5kJ always produces 10m/s, Why does the second 5kJ only produce 4.1m/s? What is going on here?
57534
I read in  in this site that the KE a free-falling ball acquires is not originated by the attracting body but that energy was actually stored in the ball when it had been lifted to the height it dropped from.  In this way, it was said, gravity is subject to the conservation of energy principle and cannot change the total energy of an object.  Consider now the maneuver known as a  (also gravity assist) used by space probes such as the . A space probe approaches a planet with velocity $v$, slingshots around, and ends up with velocity $v+2U$, where $U$ is the velocity of the planet.  Consider the energy of the probe. Before, it was $E_i=\frac{1}{2} mv^2$ and after it is $E_f=\frac{1}{2} m(v+2U)^2$. It looks like $E_f$ is much bigger than $E_i$ - but where did the additional energy come from?   Is this not a violation of the conservation of energy principle?
109368
So, in my physics textbook it's written that the kinetic energy possessed by a body is equal to the amount of work done by an opposing force to stop the body, now that got me thinking that when a ball hits a wall, the wall which is the opposing force does no work  because there is no displacement, so to say isn't the kinetic energy possessed by the ball zero?
31834
I don't understand why it is said that the (rest) mass of a system is not conserved in relativity. I mean, the momentum of a system is conserved (i.e.: it remains constant in a frame of reference without any external influence). Also the energy of a system is conserved in relativity: it doesn't change without any external agency  and the (rest) mass of a system is just  $$m^2=E^2-p^2$$  $E$ is constant, $p$ is constant, how can $m$ change?  For example, entropy is not conserved in a system, that means that the entropy of the system will increase spontaneously with time, which is really the case. But, is this true for mass?   here is the problem from Griffiths and exact solution as given in the Griffiths,  Two lumps of clay, each of rest mass $m$, collide head-on at $3c/5$ and they stick together. Question: what is the rest mass $M$ of the composite lump?  Solution: In this case conservation of momentum is trivial it is zero before and zero after. The energy of each lump prior to the collision is $$mc^2/\sqrt{1-v^2/c^2} =5mc^2/4.$$ The energy of the composite lump after the collision is $Mc^2$ (since its at rest). so the conservation of energy says: $5mc^2/4 + 5mc^2/4 =Mc^2$ and hence $M=5m/2$.  Notice that this is greater than the sum of the initial masses! Mass was not conserved in this collision; kinetic energy was converted into rest energy, so the rest mass increased.
78725
Kind of an odd, random question that popped into my head. Tidal energy - earth's ocean movement, volcanism on some of Jupiter's moons, etc. - obviously comes from the gravitational interaction between large bodies. On earth the interactions with the moon are pulling water around the surface, creating some amount of heat due to friction, etc.   My question is, where does that energy come from exactly? More specifically, what potential energy source is getting depleted to do that work? Is the earth minutely slowing down in its spin - or are the orbits of earth and the moon subtly altered over time by the counteractive movement and friction of liquids and gasses?
40049
I would like to know where gravity gets its energy to attract physical bodies?  I know that the law of conservation states that total energy of an isolated system cannot change. So gravity has to be getting its energy from somewhere, or else things like hydropower plants wouldn't be able to turn the power of the falling water into a spinning rotor.  Just to be clear, Lets create an example:  Lets say we have two objects with equal mass close to each other. So gravity does its job and it pulls each other closer, this gets turned into kinetic energy. This is where I'm lost. According to the law of conservation energy can't be created or destroyed and the kinetic energy comes from the gravitational pull so where does the gravitational pull gets its energy.  If that energy isn't being recycled from some where else then that means you have just created energy, therefore breaking the law of conservation.
43205
I'm a physics tutor for 11th and 12th grade students. I taught my students that momentum is conserved only when the net external force on the system is zero. But in the books we are following, momentum was conserved in the following cases in spite of presence of an external force:   A body (mass m1) is dropped from a tower (height h) and another body (mass m2) is projected vertically upward with initial velocity u. Find the velocities just after collision. (Gravitational force is the external force) Two bodies projected towards each other on a rough table surface, with all required parameters given asked to find the velocities after collision. (frictional force is the external force) Body explodes in mid air. Momentum is conserved before and after explosion. (Again gravitational force is the external force)   I explained to my students in all these cases though an external force is present it has negligible effect. But I'm not satisfied with my answer.  EDIT: Thank you for the comments and answer. It is clear that contribution of external force to change in momentum is negligible. Can someone do the calculation of taking change in momentum into consideration and showing that it is negligible?
14508
This question has been asked many times all over the Internet and answers can be found on places such as yahoo and ask.com, but I'm not satisfied with those answers and I don't trust the validity of those places.  This seems like a more appropriate place to ask.  So, here's my train of thought and I would like to ask you if you think I'm right or wrong.  Even though most of the answers point to the fact that the ball does not achieve equilibrium because the force of gravity is constantly acting upon it, thus causing an acceleration, I still think there's a moment of equilibrium and here's why:  When the ball is traveling up, it's accelerating towards the ground and eventually reaches a point at which its speed reaches 0.  At this moment, isn't the reason the speed is 0 is due to the fact that net force is 0?  The force that made the ball travel in the upward direction was canceled out by the gravitational force.  So, the sum of forces at that brief moment is equal to 0, otherwise the ball would be moving.  I don't think gravity is the only force acting on the ball at that moment, since the throwing force was acting on it as well, until both canceled each other out for that brief moment.  Am I right or wrong?
22605
Sorry to ask the long way, but trying to make the question clear so that I can get a clear answer.   Why does it make any physical sense for a body to have negative potential energy?   If a body A is at infinite distance from another body B, it has zero gravitational potential energy, but if it is at "zero" distance from another body B, it has negative infinite potential energy, and acquires infinite kinetic energy in the process of getting there.  Please don't tell me that the zero point of potential energy is arbitrary, and it could just as well have zero potential energy when it is at zero distance from another body B, and positive infinite potential energy when it is at infinite distance from the other body B. How can it have this? When body A is at infinite distance from the other body B, then it is not influenced by the other body B at all, it is as if the other body B does not exist at all, so then the potential energy in body A must be zero, as there is no gravitational field of the other body B to cause any potential energy in body A.  But then the question is if the potential energy contained in the body A is zero at infinite distance from the other body B, then how is it possible for this zero potential energy to be reduced further and get changed to kinetic energy as the body A accelerates towards the other body B, so that the potential energy becomes negative? As I see it, if the body A never had any potential energy to begin with, then this not present energy cannot change into kinetic energy, otherwise we would be creating kinetic energy out of nothing.   Why does it make any physical sense to say that the body is in "potential energy debt", ie the potential energy is reduced to negative when it was zero to begin with?
74069
Almost in every book on physics, there's an example of conservation of momentum when the ball that is moving horizontally in the air, hits some massive wall. They claim that the return speed of the ball when it bounces off is the same as it was before the hit. If there were no external forces acting on the system (or their net force was zero) that would be fine. But in this case, there is a gravitational force acting on the ball, and because there is no surface underneath it, there's no normal force and therefore it doesn't "cancel out" the gravitational force. So my question is, why they say that the momentum conserved? Do they neglect the gravitational force or what? I'm quite confused.