Page 1Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 1Lecture 14Goals:Goals:Assignment: Assignment: l HW6 due Tuesday Oct. 25thl For Monday: Read Ch. 11••Chapter 10 Chapter 10 v Understand the relationship between motion and energyv Define Kinetic Energyv Define Potential Energyv Define Mechanical Energyv Exploit Conservation of energy principle in problem solvingv Understand Hooke’s Law spring potential energiesv Useenergy diagramsenergy diagramsPhysics 207: Lecture 14, Pg 2Kinetic & Potential energiesl Kinetic energy, K = ½ mv2, is defined to be the large scale collective motion of one or a set of massesl Potential energy, U, is defined to be the “hidden”energy in an object which, in principle, can be converted back to kinetic energyl Mechanical energy, EMech, is defined to be the sum of U and Kl Others forms of energy can be constructedPage 2Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 3Recall if a constant force over time thenl y(t) = yi+ vyit + ½ ayt2l v(t) = vyi+ aytEliminating t givesl 2 ay( y- yi) = vx2 - vyi2l 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 right½ m vyi2+ mgyi= ½ m vyf2 + mgyfWe now define mgy U as the “gravitational potential energy”A relationship between y- displacement and change in the y-speed squaredPage 3Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 5Energy (throwing a ball)l Notice that if we only consider gravity as the external force then the x and z velocities remain constant l To ½ m vyi2+ mgyi= ½ m vyf2 + mgyfl Add ½ m vxi2+ ½ m vzi2and ½ m vxf2 + ½ m vzf2½ m vi2+ mgyi= ½ m vf2 + mgyfl where vi2 vxi2 +vyi2 + vzi2½ m v2K terms are defined to be kinetic energies(A scalar quantity of motion) Physics 207: Lecture 14, Pg 6When is mechanical energy not conservedl Mechanical 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.Page 4Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 7Inelastic collision in 1-D: Example 1l A 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.v What is the initial energy of the system ?v What is the final energy of the system ?v 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 ?v What is the initial energy of the system ?v What is the final energy of the system ?v Is momentum conserved (yes)? v Is energy conserved? Examine Ebefore-EaftervVbefore afterxvrm v212m V)(212Mm + V)( 0 M vMmm+=+)(1v21 vv)(21 v21 V]V)[(21 v21222MmmmMmmmmMmm+−=+−=+−No!Page 5Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 9Elastic vs. Inelastic Collisionsl A collision is said to be inelastic when “mechanical” energy ( K +U ) is not conserved before and after the collision.l How, if no net Force then momentum will be conserved. Kbefore+ U ≠≠≠≠ Kafter+ U v E.g. car crashes on ice: Collisions where objects stick togetherl A collision is said to be perfectly elastic when both energy & momentumare conserved before and after the collision. Kbefore+ U = Kafter+ U v Carts colliding with a perfect spring, billiard balls, etc.Physics 207: Lecture 14, Pg 10EnergyllIf only If only ““conservativeconservative””forces are present, then theforces are present, then themechanical energy of a systemmechanical energy of a systemis conservedis conservedFor an object acted on by gravity K and U may change, K + U remains a fixed value.Emech= K + U = constantEmechis called “mechanical energy”½ m vyi2+ mgyi= ½ m vyf2 + mgyfPage 6Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 11Example of a conservative system: The simple pendulum.l Suppose we release a mass m from rest a distance h1above its lowest possible point.v What is the maximum speed of the mass and where does this happen ?v To what height h2does it rise on the other side ?vh1h2mPhysics 207: Lecture 14, Pg 12Example: The simple pendulum.yy=0y=h1v 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.Page 7Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 13Example: The simple pendulum.vh1yy=h1y=0v 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 = mgh1at topE = mgh1= ½ mv2at bottom of the swingPhysics 207: Lecture 14, Pg 14Example: The simple pendulum.yy=h1=h2y=0To what height h2does 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 = h2Page 8Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 15Potential Energy, Energy Transfer and Pathl A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless1. 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 … againl To complete the loop the loop, how high do we have to let the release the car?l Condition for completing the loop the loop: Circular motion at the top of the loop (ac= v2 / R)l Exploit the fact that E = U + K = constant ! (frictionless)(A) 2R (B) 3R (C) 5/2 R (D) 23/2Rh ?RCar has mass mRecall that “g” is the source of the centripetal acceleration and N just goes to zero is the limiting case.Also recall the minimum speed at the top isgR=vUb=mghU=mg2Ry=0U=0Page 9Physics 207 – Lecture 14Physics 207: Lecture 14, Pg 17ExampleThe Loop-the-Loop … againl Use E = K + U = constantl mgh + 0 = mg 2R + ½ mv2 mgh = mg 2R + ½ mgR = 5/2 mgRh = 5/2 RRgR=vh ?Physics 207: Lecture 14, Pg 19Variable force devices: Hooke’s Law Springsl Springs are everywhere, l The magnitude of the force increases as the spring is further compressed (a displacement). l Hooke’s Law,Fs= - k ∆s∆s is the amount the spring is stretched or compressed from it
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