Potential Energy as a Funxtion of x?
813028
Well, as you've written it, it's a matter of differentiation the first equation, which gives you the second, as long as you take the limits of the integral properly.\n\nObjects in a potential field have an energy due to that field, dependent on their position, x. They will tend to move in a direction that lowers their energy, and experience a force due to the field that pushes them in the right direction. Hence, after some maths, the second equation - the force experienced due to a potential field is equal to the rate of change of that potential with respect to position - if the potential field changes rapidly, the force will be stronger. The minus sign means that the force moves you from high energy to low energy, ie, U decreases as the force moves you.\n\nThe first equation states much the same thing, the change in potential energy is from the force acting over a distance. This is usually how potential energy is defined, in case you're wondering where is came from.\n\nWrite Delta u as U(0) - U(x): the change in energy, moving from your starting point x=0 to some other point. Then the integral is\n\nS F(x) dx, with upper limit of x and lower limit of 0.\n\nDifferentiate both sides. The "U" side gives you -dU/dx, because U(0) is a constant, and the "F" side gives you F(x).\n\nIt's hard to explain any further without diagrams and stuff. Most first year college physics texts have more detailed explanations, better than I can. Hope this helps.
962133
E potential = mgh\n= 4 kg ( 9.81 m/s^2) ( 2.5)\n=98.1\n\nPotential energy was not lost....it transformed into Kinetic Energy\n\nKinetic + Potential = Total Energy\n\nLaw of Conservation of Energy
150385
f(x)=Int(f'(y), 0, x)-f(0) => yes, they're opposite operation, but if x has a given value, e.g. x=2, 3...no, they're not, b/c in that case, the RHS of the previous egality is a real number, not a function
950482
OK, Drummanmatthew.  It's been a long time since I did this but I think I remember enough to help you out.\n\nFrom what I remember, the potential, phi, is a function of x, y and z defined so that F = -grad(phi)  (I'm not clever enough to do symbols, I'm afraid) where grad is the operator (d/dx, d/dy, d/dz) where the d's are actually partial differential operators, rather than ordinary differential operators.\n\nIn other words, we need a function which when you diff wrt x gives you -2xy, when you diff wrt y gives -x^2-z^2 and when you diff wrtz gives you -2yz.  To me, it is the y part of this that gives it away.  It clearly has to be something plus something.  You need the bit with the z to differentiate away when you diff wrt x.  It seems to me that -yz^2 will do the trick.  Diff wrt x and it goes, diff wrt y and you get -z^2.  Diff wrt z and you get -2yz, as required.  To get the rest of it right you also need a -xy^2 by similar logic.\n\nSo, I reckon that phi = -y(x^2 + z^2) + c\n\nThe constant c should be obtainable from a boundary condition somewhere (I seem to recall that we often want phi to go to 0 as we go off to infinity but this is clearly not possible here).\n\nI'm not absolutely certain about the - but I think this correct apart from that.\n\nHope this helps.
958996
initial energy : (mv^2)/2 = (0.047*56^2)/2 = 73.696 J\nfinal energy : mgh +KE= 0.047*9.81*17 + KE = 73.696\n\nKE = 73.696 - 7.83819 = 65.85781 = (mv^2)/2\nv = squareroot(KE*2/m)\nv= 52.94 m/s
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
352249
This integral does exist (it can be evaluated for any endpoints you choose). Since it cannot be represented as a finite combination of elementary functions and operations, it is given its own name, the error function. Many integrals like this have their own names and are quite useful in applied mathematics. See http://mathworld.wolfram.com/Erf.html .\nTo say that it does not exist is unrigorous.
443622
For element X the amount of energy required to remove one of the the outermost electrons to make the ion X+ (hence the name Ionisation)
345069
This can be shown for X as either continuous or discrete.  Let mu = E(X).\n\nVar(Y) = E(Y^2) - E(Y)^2\n= E[(X-mu)^2] - E(X-mu)^2\n= E(X^2 - 2Xmu + mu^2) - (E(X)-mu)^2\n= E(X^2) - 2muE(X) + mu^2 - (mu-mu)^2\n= E(X^2) - 2mu^2 + mu^2 - 0\n= E(X^2) - mu^2\n= E(X^2) - E(X)^2\n= Var(X)\n\nIf you need to be more specific than this, the replace E(X) = int(-inf to inf) xf(x)dx and E(X^2) = int(-inf to inf) x^2f(x)dx.  Everything should come out the same way.\n\nRemember that using substitution in the integral, let x = y + mu, the dx = dy, and the limits of integration would remain -inf to inf.
1037424
evenly distributed energy
348897
Yes, the answer is B.\n\nThere are two ways to do this:\n\n1. Use the law of conservation of energy:\n\nKE = PE\n\n1/2 mv^2 = mgh\n\nso:\n\n1/2 v^2 = gh\n\nso:\n\nv^2 = 800\n\ngiving v = roughly 28.3 thus B\n\n2. Use kinematic equations:\n\nvf^2 = vi^2 + 2ax\n\n0 = v^2 + 2(40)(-10)\n\nv^2 = 800 giving v = roughly 28.3 thus B\n\n:)
926088
So the first one can be solved easily using the energy method. Potential energy = mgh = 2*9.81*6\nKenetic energy  = 1/2mv^2 = 1/2*2 * v^2\nBecause there is no friction Potential Energy = Kenetic Energy:\n\n2(9.81)(6)=1/2(2)(v^2) \n\nNow you just need to solve for v, using your calculator or head.\n\nFor the second one you have to account for the amount of energy lost from friction. So, instead of KE=PE you get:\n\nKenetic Energy = Potential Energy - Energy lost by Friction\n1/2*m*v^2 = m*g*h - Friction energy \n\nSo you can do some algebra and\nm*g*h - 1/2*m*v^2 = Energy lost by friction \n\n\nThree is impossible without more information. You can't solve for u with just m, theta, and work. You need some information about the height, or the length of the ramp, or end velocity. Just think about it, you could have a ramp 100ft tall with little friction, or you could have one 1 ft tall with lots of friction. The work done would be the same even if you kept the masses and theta the same.
715637
Work and energy always involves comparing energy before and after\n\nKE[before] + GPE [before] = KE[after]+GPE[after]\n\n(KE=kinetic energy; GPE=gravitational potential energy)\n\nor more advanced situtations include EPE (elastic potential energy) but don't worry about this.\n\nYou should know that KE = 1/2mv^2 and that GPE=mgh\n\nUsing the inclination and angle should get you the height. \n\nUsing F=μR and F=ma and manipulation, involving replacing a with speed, should get you the answer.\n\nAlways draw a diagram and resolve horizontally and vertically\n\nHope this helps
916732
Is relative to the amount of energy and object contains.
711054
It's two.  Energy goes as the square of amplitude.
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
738204
Wouldnt that just be Work=Work\n=> m x g x h = m x (lambda)\n=> gh=(lam) ?