Potential Energy, Kinetic Energy (and Potential Forces).  What is the speed of the object at point A?
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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:)
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The equations of motion you are using are good equations, but aren't going to answer your question.  What you want to know is the energy of the moving object.  \n\nThe kinetic energy of a moving object is E = (1/2)m*v^2.  From this equation you can see that if you double the velocity, the energy quadruples.  Stopping an object with four times the kinetic energy will take 4 times the distance.
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Kinetic energy.\n\nGreek kntikos, from kntos, moving, from knein, to move.
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The unit of force is the Newton (N).\n\n1 newton is defined as the force needed to accelerate a 1kg object at 1m/s*2.(from Newton's 2nd Law of Motion: F = ma)   \n\nSometimes it is given in its base units as kgm/s*2
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zero, \n\nno acceraltion = no net force\n\nmathematically  Net F = ma\n\nif the right side is 0, left side is too
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a)F = MV=  m1v1 +/-  m2v2  \nusing these we get,  95*4.1 + 85*5.5 (since same direction)=857kgm/s = (95+85)* V. so, V = 4.76m/s\nb) MV = M (6) = (9500+m)*6 = 9500*16 + m*0. \nso, (9500+m) = 9500*16/6= 25333.33 kg\nso, m = 25333.33-9500 =15833.33kg\nc)work done against friction= kinetic energy of bullet\nW = FScos(tetha)  since its horizontal, angle is zero, \nand cos0 = 1, so W = FS\nwe know, coeff of friction = tangent of angle between, the frictional force  and normal reaction...\nso, the \nF = normal reaction=(1.1 +0.015)* 9.81=10.938N\nfrictional force= 0.25 *10.938 = 2.7345N\ndistance is 9.5m...\nso W = 2.7345*9.5 = 25.97 J\nthe kinetic energy of bullet has been transferred to the block...\nso the kinetic energy of bullet is 25.97J\n1/2*(0.015)* v^2 = 25.97\nso, v^2 = 25.97*2/0.015=3463.74\nso, v = 58.85m/s\nhope this helps...:)
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Kinetic Energy = 1/2mv^2; Potential Energy = mgh\n\n1/2mv^2 = mgh or 1/2v^2 = gh\n\nv^2 = 2gh = 2 * 9.81m/s^2 * 0.49 m\n\nv = 3.1 m/s
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Please take the time to\nvote for the best answer\n\nThank You!!\n\nRecall that an object of a certain mass moving with particular speed will have an associated kinetic energy  mass x speed2. An object spinning about an axis will also have associated with it a kinetic energy, composed of the kinetic energies of each individual part of the object. These individual contributions may be summed up to give an expression for the total kinetic energy of the spinning object: \n\nLook this up for more
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Assuming that the energy of the ball remains constant (i.e. no air resistance), the speed will be the same. The VELOCITY, however, will be -35 m/s.\nSpeed and velocity are NOT the same!
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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?
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1 sum the forces and set equal to zero. 2 forces, gravity and electric.\n\n2. conservation of energy. initially all kinetic, finally all potential. KE1=U2. U2 is electric potential. solve.\n\ngood luck
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Energy can go from place A to place B (think about sound and light)\nEnergy can go from object A to object B (think about a collision)\n\nTh
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Kinetic Energy = mass x velocity^2 divided by 2\n\nKE = 12 x 12^2 / 2
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F = d/dt(mv) [the time deravitive of the momentum] = dp/dt  where p is the momentum\n\nWork = Energy = F x distance\n\nPower = Energy/time\n\nPower = (F x distance)/time\n\nPower = [(dp/dt) x distance]/t\n\nremember dp/dt = d/dt(mv) = m dv/dt + v dm/dt
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DEFINITION:\n\nKinetic energy of a moving body is defined as the work done to bring it to rest. \n\n\nILLUSTRATION: \n\nIf a particle of mass 'm' is moving with velocity 'u' along a straight line and if a force 'F' is applied (in a direction opposite to its velocity) on it to bring it to rest after travelling 's' distance then the work done to bring the body to rest from its present state of motion is '-F*s' (force*displacement; - sign appears as work is done on the body and not by the body). \n\nWe know from mechanics 'V^2=u^2+2*f*s'_______(1) \nwhere 'v'=final velocity, 'u'=initial velocity and 'f'=constant acceleration of a particle in rectilinear motion (i.e. motion in a straight line).\n\nHere final velocity is zero as the body comes to rest finally, so v=0.\n\nAlso, f=F/m (since from Newton's second law F=m*f).\n\nSubstituting these values of 'v' and 'f' in eqn. (1) we get\n0 = u^2+2*(F/m)*s\nor s=-(m*u^2)/(2*F)___________(2)\n\nThus the work done to bring the body to rest from its present state of motion = -F*s = (-F)*{-(m*u^2)/(2*F)} = (m*u^2)/2. This is the expression for kinetic energy of a body in translation.\n\nUNIT:\n\nS.I. unit: Joule\nM.K.S. unit: erg\nN.B.: 1Joule=10^7erg
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Potential energy is the energy by virtue of its position.\nAny object which is far away from the center ofpossessingf nothing prevents it will move towards the center of earth. A spring wound fully  or partially. If left alone and it will rewind to its normal position.A  vessel containing gas under high pressure.If there is a vent the pressure will drop to normal.A string under tension. All thesrelievedamples of potential energy. These energies will be releived from the object when  there is proper way. \n\nKinetic energy is the energy by virtue of its motion. Any object moving in other words having velocity is possesing kinetic energy.If the velocity is reduced by some external agency the latter gets energy and viceversa.
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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.