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Motion, Forces and Energy Lecture 7: Potential Energy & Conservation

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 U g = m g h

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Motion, Forces and Energy Lecture 7: Potential Energy & Conservation

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  1. Motion, Forces and EnergyLecture 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

  2. Conservation of Energy Total Energy, E1 h1 v1 y h2 v2 Total Energy, E2 The change in total energy is DE where: 0

  3. 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: yi = h Ui = mgh KEi = ½ mvi2 yf = y Uf = mgy KEf = ½ mvf2 h y y=0 Ug=0

  4. 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 But Ei = Ef due to energy losses due to friction so energy loss DE is d=1.0m DE = -fk.d = -mk mgcosq.d = -5.0J So ½ mvf2 = 14.7-5.0 J And therefore vf=2.54 ms-1 h=0.5m vf 30o

  5. m1=3.0 kg fk m2=5.0 kg 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 Change in PE for the 5 kg mass is DU5kg = -m2 g d = -73.5 J Without friction, the KE gained by BOTH masses would sum to DU5kg (73.5 J), but With friction, the KE gained = 73.5 – 17.6 = 55.9 J DKE = ½ (m1+m2)v2 (both masses move with same velocity) So v = 3.7 ms-1.

  6. 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 Vertical height travelled by the mass = (d+x)sinq=h Change in PE of the mass, DUg = 0 – mgh = -mg(d+x)sinq Elastic PE of spring increases from zero (unstressed) to: DUs = ½ kx2 So DUg + DUs = 0 So mg (d+x) sinq = ½ kx2 giving d as: d k q

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