Physics --- Work and Energy --- Please help!!!!!!!!!!!!!!!!!!?
913728
1. KE = (1/2)*m*v^2 - you have the KE and mass.\n\n2. The amount of work done will be equal to the kinetic energy the arrow is given. \nW = F*d = (1/2)*m*v^2 solve for v.\n\n3. When the speed is increased 50% Vnew = 1.5*Vold. Kinetic energy will increase by 1.5^2 = 2.25. \n\nWork done stopping it is Fbrakes*distance and equals the kinetic energy it has. \nKEnew = 2.25*KEold so the stopping distance, SD, changes accordingly:\nSDnew = 2.25*SDold\n\n4. The acceleration on the baseball is given by\nV^2 = Vo^2 + 2*a*d where V is final velocity, Vo is original velocity, a is acceleration, and d is distance. Finally, \nF = m*a.\n\n5. \n(a) The tension is divided between holding the weight and accelerating it. T = W + m*a.\n(b) The work gives the load a potential energy. \nU = m*g*y\nIt also gives it kinetic energy. The velocity it has at the 21.0 m height is given by\nV^2 = Vo^2 + 2*a*y. Use the velocity in the KE equation. The work done is the sum U + KE.\n(c) Might be a trick, but this feels like semantics. I don't see that \nb & c are any different. Maybe in class, or the book, some distinction was made between those 2 ways to say it?
1105028
read this\n\nhttp://en.wikipedia.org/wiki/Energy
47074
If the force acting on the object doesnt result in displacement of the particles in relation to the surroundings of the system under consideration, then the work done is zero.\n\n  Mathematically the work done is zero when the displacement is at right angle (90 degrees) to the direction of the force. \nSince work is calculated as force x displacement x cos(theta) \ni.e. W = F x d x cos (theta) \nwhere theta is the angle between the force and the displacement.\n\nEg : The earth does no work on the moon since the motion of the moon is at right angle (tangential) to the direction of the force applied by the earth.
922769
Okay so start off by looking at the definition of work.\n\n"Work is a force exerted on an object that causes it to have a displacement in the same direction.\n\nW = Fd\n\nW = work (Joules), F = force (Newtons), d = displacement (metres)\n\nBy definition, 1 J of work is done by applying 1 N of force to move an object 1 m.\n\n    * Work has no direction, so it is a scalar quantity.\n    * Also, the direction of motion must be in the same direction (parallel) as the direction of the force, otherwise no work is being done."1\n\nSo there are two things you could look at here.\n\nW = Fd\n\nHow much displacement is there in one full cycle of something moving in a circle?  Displacement is the change in position so look at the position at the beginning and ending of the cycle.  There is no change in position so d=0 which gives\n\nW=F(0)\nW=0.\n\nNo work done.\n\nAnother thing you have to look at is "Also, the direction of motion must be in the same direction (parallel) as the direction of the force, otherwise no work is being done"\n\nIn uniform circular motion the movement of the object at any point in time is tangential to the circle while the force is always directly towards the centre of the circle so the movement and force are never in the same direction.  The definition of work demands that they be in the same direction for any work to have been done and, since they are not, no work has been done.\n\nShort answer: c. 0 J
696257
Energy given = energy absorbed\ntotal travel of pile driver h1 =4.00+0.11=4.11 m\nAssume the pile has no mass\nenergy given = mgh1 = 1900*9.81*4.11\nenergy absorbed= F * h2 = F * 0.11\nF * 0.11 = 1900 * 9.81 * 4.11\nF=696,420.82 N
1128176
work = change of kinetic enegy\nW = 1/2 m (vf² - vi²)\nW = 0.5 * (0.250 kg) * (16² - 20²) m²/s²\nW = -18 J
884398
The work done on the car by the net force equals the change on the car's kinetic energy.\n\nWork = Force * distance\nKE = 1/2 mv^2\n\nWe are given a mass for the car and the final velocity, so from this we can calculate the car's final KE,\nKE = 1/2 (2000 kg) * (2 m/s)^2\nKE = 4000 Joules\n\nSince in this case, KE = W\nW = 4000 Joules = F * d\n\nWe know the net force to be the sum of all the forces acting on th car.  Since there is a "forward" force of 1140 Newtons and a "backward" (what we will call 'negative') force acting on the car of 950 N, the net force is 1140 - 950 = 190 Newtons\n\nW = 4000 J = (190 N) * d\nd = 4000 J / 190 N\nd = 21.05 meters
310103
Kinetic Energy:\n\nKinetic energy (SI unit: the joule) is energy that a body possesses as a result of its motion. It is formally defined as the work needed to accelerate a body from rest to its current velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. Negative work of the same magnitude would be required to return the body to a state of rest from that speed.\n\nIn simple cases, this work (and thus the kinetic enrgy) is equal to:\n\n    Ek =.5(M*(V^2))\nwhere m is the object's mass and v is the object's velocity.
666315
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
796831
Ei - A= Ef (A = work done by air resistance, "i" means initial and "f" means final)\n\nPEi + KEi - A = PEf + KEf\n\nA = PEi + KEi - PEf - KEf\n\nA = mghi + (1/2)mv²i - mghf - (1/2)mv²f\n\nA = m[ghi + (1/2)v²i - ghf - (1/2)v²f]\n\nA = 0.664[(9.80)(2.27) + (1/2)(7.00²) - (9.80)(3.31) - (1/2)(5.23²)]\n\nA = 0.664[22.2 + 24.5 - 32.4 - 13.7]\n\nA = (0.664)(0.6)\n\nA = - 0.4 J (since the work done decreases the total energy, it's negative)\n\n*** 0.4 J of work are done BY the nonconservative force air resistance ON the ball.\n\n\n*** Believe it or not, the engineer above me did it completely wrong, even though he understood the concept.  First, work and energy are NEVER expressed in Netwons (N), but in other units such as joules (J).  Also, when calculating the potential energy, the formula is PE = mgh.  The engineer forgot the "g," which is equal to 9.80 m/s^2.  I am fresh out of AP Physics, and I got a 5 on the AP test.  All throughout physics and AP physics, I was drilled with all kinds of problems.  Energy/work problems are my specialty.  You can even verify my answer with your friends and your teacher, and you'll see that it is correct, and the engineer's answer is very wrong.\n\n*** To the engineer: Sometimes, even professionals can be wrong.  I'm sure you understand all the concepts of physics.  You obviously have done a LOT of math and physics in the past, since you're an engineer.  However, it's probably been a while, and you may have forgotten a few formulas (and I don't think the potential energy formula can be easily derived).
987930
No energy source? What source of energy is needed? And the simple presence of mass causes gravitons. And these gravitons interact with one another. And proof, you need proof that gravity works? Dude go jump off of a building and see what happens!
302252
You are looking for the amount of work done by the bicycle.\nWe know this value will be less than 5.4X10^3 J because energy is always lost when using a machine.\nIn addition, the answer is equal to the efficiency (62%) times the amount of work the cyclist does. \n(I wouldn't be right if I did your whole homework problem so I'll stop here)
1119730
ahh, good theory problem...\nlet's see, there's energy, friction, force (normal, tension, as well as the other ones mentioned), gravitation, mass, pendulum motion, gotta be more that I've missed, but no doubt someone'll be here to correct me.
822266
Conservation of energy:  KE=PE\n1/2*Mb*Vb^2= (Mb+Mw)*g*h\nVb^2=2*(Mb+Mw)*g*h/Mb\nVb^2=2*(0.01+2)*10*0.1/0.01\nVb^2=402\nVb=20.05 m/s
249841
Work equals the vectors of energy.
961494
This is an energy balance situation.  The work done in dragging the object must equal the energy change.  In this case the energy is potential energy, given by m*g*∆h, where ∆h is the change in vertical distance.  The work (energy) required to drag the block is force times distance along the incline.  Equate these two quantities and solve for F.
2752
A. Potential energy = Force * Distance\n     = (50 * 9.8) * 10\n     = 4900 J\nB. Power rating = Energy / time = 4900 / 5 = 980 Watts