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# animation about potential energy

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Question:A scale contains a spring constant of 275 N/m. Placing a mass on the scale causes the spring to be compressed by 3.25 cm. Calculate the elastic potential energy stored in the spring.

Answers:Potential Energy = kx where k = spring constant = 275 N/m x = distance travelled = 3.25 cm = 0.0325 m PE = 275 * (0.0325) => PE = 0.290 J

Question:Is it only in terms of gravitational and electromagnetic? Or does it encompass potential energy in terms of strong and weak nuclear forces too? And how about complete conversion to energy by matter antimatter reaction? Is that a potential energy too?

Answers:Basically, potential energy is stored energy that is waiting to be used. The opposite of potential energy is kinetic energy, energy that involves motion. But one thing that you should remember is that every single object existing has energy no matter if its moving, still, hot, cold, spherical or a cube. Everything has energy.

Question:Sorry, I'm REAL bad with these sorts of problems... For a particular hunting bow, the spring constant for the bow is 700 N/m. The arrow has a mass of 0.91 kg. How far must the string be drawn back to give the arrow an initial speed of 50 m/s? If the arrow is aimed straight up, how high will it reach? Any help is greatly appreciated. =)

Answers:Conservation of Energy: Initial Energy = Final Energy Elastic PEi + KEi = Elastic PEf + KEf Elastic PE = 1/2kx^2 KE = 1/2mv^2 PEf = 0, KEi = 0 1/2(200N/m)(x)^2 = 1/2(0.91 kg)* (50m/s)^2 Solve for x

Question:I have just done a problem about a box sliding on an inclined plane with friction. The question was done from an energy perspective but I can't understand why the component of weight down the plane i.e mgsin(o) is ignored. Also in force and potential energy what does it mean to talk about the "potential energy of the system" i.e when talkin about several charged particles. Finally I realise that force is equal to the negative of the derivative of potential but am struggling to get my head around the statement when F(x) is in the positive direction U(x) decreases with increasing x. Can anyone think of a simple example to show this in action. I probably can't because it relates to the "potential energy of the system" (above) which I don't properly understand. Thanks in advance for any answers.

Answers:Part 1) The component of the force that points in the x direction of the does not configure into the problem because it has no bearing on the energy of the system. There is potential energy at the top of the incline and energy lost due to friction. When dealing with energies you look at the initial and final energy states of the system. If you have velocity at the end then that can be related to mgsin(o) if you want to think of some connection. When dealing with energies you have: Potential Energy: mgh (for objects at height h), 1/2 kx^2 (for compressed or elongated springs) Kinetic Energy: 1/2 mv^2 (object in motion) Energy loss due to friction: f*d (friction force over distance) You can find v using your equation of motion and force equations that deals with mgsin(o) if that is the connection you are looking for. Part 2) Charged particles have potential energy between them like massive objects have gravitational potential energy between them. When dealing with objects with mass the newtonian gravitational relationship is given by: F = -GMm/r^2. This is what you use to figure out the potential energy etc. For charged particles the coulomb force is related similarly by this equation: F = kQq/r^2. The Q's are the charges on each particle similar to the masses in the gravity equation. k is just a constant. Notice how both equations have the inverse square distance relationship. From this you can conclude that charged particles have potential just like massive objects. I find it strange that one is dealing with E & M during a mechanics course though! Part 3: I do not think you should be thinking of the negative as a direction. That will cause you much confusion. Remember that gravitational forces are always attractive forces and not repulsive. That is why the equation is set up the way it is. From the force equation you see above for gravity is negative. That means the force is attractive and it does not mean that the force is moving in the negative x direction. If you are looking at two objects you can always orient the objects along a straight line and the negative will imply that the force is pushing/pulling one object towards another. This might help you understand the signs a bit. It is confusing because sometimes the equations you get do imply a direction along the axis and sometimes it does not depending on the equation you have. Hope that helps.