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Motion, Forces and Energy Lecture 7: Potential Energy & Conservation The name potential energy implies that the object in question has the capability of either gaining kinetic energy or doing work on some other object: Gravitational Potential Energy Ug = m g h Elastic Potential Energy Us = ½ k x2 Conservation of Energy h1 v1 Total Energy, E1 v2 Total Energy, E2 y h2 0 The change in total energy is E where: 1 1 2 2 E E2 E1 mgh2 mv2 mgh1 mv1 2 2 1 2 2 E mg h2 h1 m v2 v1 2 2 2 2 2 As v2 v1 2 gy, then v1 v2 2 gy 1 So E mgy m 2 gy 0 2 Object in free fall A ball of mass m is dropped from a height h above the ground. We can use energy calculations to find the speed of the ball when it is either released from rest or with an initial speed vi: Generally : KEi U i KE f U f yi = h Ui = mgh KEi = ½ mvi2 yf = y Uf = mgy KEf = ½ mvf2 h y 1 1 2 2 mvi mgh mv f mgy 2 2 2 2 v f vi 2 g (h y ) v f vi 2 g (h y ) 2 From rest : v f 2 g (h y ) y=0 Ug=0 Energy losses (non-conservative forces) Kinetic friction is an example of a non-conservative force. When a book slides across a surface which is not frictionless, it will eventually stop. But all the KE of the book is NOT transferred to internal energy of the book. We can find the speed of the mass sliding down the ramp below if we know the frictional force by considering kinetic and potential energies: Initial energy, Ei = KEi+Ui = mgyi=14.7J Final energy, Ef = KEf+Uf = ½ mvf2 vi=0 d=1.0m But Ei = Ef due to energy losses due to friction so energy loss E is E = -fk.d = -mk mgcosq.d = -5.0J h=0.5m So ½ mvf2 = 14.7-5.0 30o vf J And therefore vf=2.54 ms-1 Another similar problem: The friction coefficient between the 3 kg block and the surface is 0.4 . If the system starts from rest, calculate the speed of the 5 kg ball when it has fallen a distance of 1.5 m. Mechanical energy loss is –fk.d = -mk m1g d (work done by friction on block) = -17.6 J m1=3.0 kg Change in PE for the 5 kg mass is U5kg = -m2 g d = -73.5 J fk Without friction, the KE gained by BOTH masses would sum to U5kg (73.5 J), but With friction, the KE gained = 73.5 – 17.6 = 55.9 J m2=5.0 kg KE = ½ (m1+m2)v2 (both masses move with same velocity) So v = 3.7 ms-1. An inclined spring A mass m starts from rest and slides a distance d down a frictionless incline. It strikes an unstressed spring (negligible mass) and slides a further distance x (compressing the spring which has a force constant, k). Find the initial separation of the mass and the end of the spring. m d Vertical height travelled by the mass = (d+x)sinq=h Change in PE of the mass, Ug = 0 – mgh = -mg(d+x)sinq Elastic PE of spring increases from zero (unstressed) to: Us = ½ kx2 k q So Ug + Us = 0 So mg (d+x) sinq = ½ kx2 giving d as: kx2 d x 2mg sin q