Lecture 14Kinetic & Potential energiesRecall if a constant force over time thenEnergy (dropping a ball)Energy (throwing a ball)When is mechanical energy not conservedInelastic collision in 1-D: Example 1Slide 8Elastic vs. Inelastic CollisionsEnergyExample of a conservative system: The simple pendulum.Example: The simple pendulum.Slide 13Slide 14Potential Energy, Energy Transfer and PathExample The Loop-the-Loop … againSlide 17Variable force devices: Hooke’s Law SpringsExercise Hooke’s LawF vs. Dx relation for a foot arch:Force vs. Energy for a Hooke’s Law springEnergy for a Hooke’s Law springEnergy diagramsEquilibriumComment on Energy ConservationSlide 28Physics 207: Lecture 14, Pg 1Lecture 14Goals:Goals:Assignment: Assignment: HW6 due Tuesday Oct. 25th For Monday: Read Ch. 11•Chapter 10 Chapter 10 Understand the relationship between motion and energyDefine Kinetic Energy Define Potential EnergyDefine Mechanical Energy Exploit Conservation of energy principle in problem solving Understand Hooke’s Law spring potential energies Use energy diagrams energy diagramsPhysics 207: Lecture 14, Pg 2Kinetic & Potential energiesKinetic energy, K = ½ mv2, is defined to be the large scale collective motion of one or a set of massesPotential energy, U, is defined to be the “hidden” energy in an object which, in principle, can be converted back to kinetic energyMechanical energy, EMech, is defined to be the sum of U and KOthers forms of energy can be constructedPhysics 207: Lecture 14, Pg 3Recall if a constant force over time then y(t) = yi + vyi t + ½ ay t2 v(t) = vyi + ay tEliminating t gives 2 ay ( y- yi ) = vx2 - vyi2 m ay ( y- yi ) = ½ m ( vx2 - vyi2 )Physics 207: Lecture 14, Pg 4Energy (dropping a ball) -mg yfinal – yinit) = ½ m ( vy_final2 –vy_init2 )Rearranging to give initial on the left and final on the right ½ m vyi2 + mgyi = ½ m vyf2 + mgyf We now define mgy ≡ U as the “gravitational potential energy”A relationship between y- displacement and change in the y-speed squaredPhysics 207: Lecture 14, Pg 5Energy (throwing a ball)Notice that if we only consider gravity as the external force then the x and z velocities remain constant To ½ m vyi2 + mgyi = ½ m vyf2 + mgyf Add ½ m vxi2 + ½ m vzi2 and ½ m vxf2 + ½ m vzf2 ½ m vi2 + mgyi = ½ m vf2 + mgyfwhere vi2 ≡ vxi2 +vyi2 + vzi2 ½ m v2 ≡ K terms are defined to be kinetic energies(A scalar quantity of motion)Physics 207: Lecture 14, Pg 6When is mechanical energy not conservedMechanical energy is not conserved when there is a process which can be shown to transfer energy out of a system and that energy cannot be transferred back.Physics 207: Lecture 14, Pg 7Inelastic collision in 1-D: Example 1A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. What is the initial energy of the system ? What is the final energy of the system ? Is energy conserved?vVbefore afterxPhysics 207: Lecture 14, Pg 8Inelastic collision in 1-D: Example 1What is the momentum of the bullet with speed v ? What is the initial energy of the system ? What is the final energy of the system ? Is momentum conserved (yes)? Is energy conserved? Examine Ebefore-EaftervVbefore afterxvm v212m V)(212Mm V)( 0 M v Mmm )(1v21 vv)(21 v21 V]V)[(21 v21222MmmmMmmmmMmmNo!Physics 207: Lecture 14, Pg 9Elastic vs. Inelastic CollisionsA collision is said to be inelastic when “mechanical” energy ( K +U ) is not conserved before and after the collision.How, if no net Force then momentum will be conserved. Kbefore + U Kafter + U E.g. car crashes on ice: Collisions where objects stick togetherA collision is said to be perfectly elastic when both energy & momentum are conserved before and after the collision. Kbefore + U = Kafter + U Carts colliding with a perfect spring, billiard balls, etc.Physics 207: Lecture 14, Pg 10EnergyIf only “conservative” forces are present, then theIf only “conservative” forces are present, then the mechanical energy of a systemmechanical energy of a system is conservedis conservedFor an object acted on by gravity K and U may change, K + U remains a fixed value.Emech = K + U = constant Emech is called “mechanical energy” ½ m vyi2 + mgyi = ½ m vyf2 + mgyfPhysics 207: Lecture 14, Pg 11Example of a conservative system: The simple pendulum.Suppose we release a mass m from rest a distance h1 above its lowest possible point.What is the maximum speed of the mass and where does this happen ?To what height h2 does it rise on the other side ?vh1h2mPhysics 207: Lecture 14, Pg 12Example: The simple pendulum.yy=0y=h1 What is the maximum speed of the mass and where does this happen ?E = K + U = constant and so K is maximum when U is a minimum.Physics 207: Lecture 14, Pg 13Example: The simple pendulum.vh1yy=h1y=0 What is the maximum speed of the mass and where does this happen ?E = K + U = constant and so K is maximum when U is a minimumE = mgh1 at topE = mgh1 = ½ mv2 at bottom of the swingPhysics 207: Lecture 14, Pg 14Example: The simple pendulum.yy=h1=h2y=0 To what height h2 does it rise on the other side?E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point.E = mgh1 = mgh2 or h1 = h2Physics 207: Lecture 14, Pg 15Potential Energy, Energy Transfer and PathA ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless 1. The ball is dropped2. The ball slides down a straight incline3. The ball slides down a curved inclineAfter traveling a vertical distance h, how do the three speeds compare?(A) 1 > 2 > 3 (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tellh1 32Physics 207: Lecture 14, Pg 16ExampleThe Loop-the-Loop … againTo complete the loop the loop, how high do we have to let the release the car?Condition for completing the loop the loop: Circular motion at the top of the loop (ac = v2 / R) Exploit the fact that E = U + K = constant ! (frictionless)(A) 2R (B) 3R (C) 5/2 R (D) 23/2 Rh ?RCar has mass mRecall that “g” is the source of the centripetal acceleration and N just goes to zero is
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