Copyright R. Janow – Spring 2012Physics 111 Lecture 06Work and Kinetic EnergySJ 8th Ed.: Ch. 7.1 – 7.5• Energy Overview• Systems• Work– Constant force 1D– Constant force in 3D– Dot Product (Scalar Product)– Units – Variable force 1D – General Vector 3D definition with variable force• The Work-Kinetic Energy Theorem• Gravitational Force and Examples• Variable (Spring) Force and Examples.• Power - Read7.1 Systems and Environments7.2 Work done by a Constant Force7.3 The Scalar Product of Two Vectors7.4 Work Done by a Varying Force7.5 Kinetic Energy and the Work-Kinetic Energy TheoremCopyright R. Janow – Spring 2012Work and Energy - OverviewDynamics: A powerful tool but intricate, detailed, vector-basedEnergy takes many forms, for example:• Kinetic: K = ½ m v2• Potential: gravitational (U = mg∆∆∆∆h), electrostatic, magnetic, nuclear• Thermal: Random kinetic energy kBT• Chemical: Binding energies in atoms and molecules • Mass-energy: E=mc2after 1905Example: Car accelerating  ∆∆∆∆W & ∆∆∆∆K positiveCar braking  ∆∆∆∆W & ∆∆∆∆K negativeEnergy in all forms is related ultimately to mechanical work. ntDisplacemex Force Work====Work-Kinetic Energy Theorem (an energy conservation statement) KK W initfinal−−−−====∆∆∆∆Work done ON systemWork is a scalarKinetic Energy is due to a particle’s speedDefinition (traditional): Energy is the ability to do workEnergy: • Scalar  Simpler approach applicable to answering certain questions• A powerful “conservation law”Energy can not be created or destroyed – conversions onlyCopyright R. Janow – Spring 2012SystemsSystems gain or lose energy when external forces do positive or negative work on them. Isolated systems have zero net interaction with the outside world and neither gain nor lose energyA system is a small portion of the Universe• The system may contain several objects within a volume of space,…• Parts of the system can exert forces on one another• External forces can act on parts of the system as well• System boundaries are chosen arbitrarily to simplify a problem –• - they separate the system from it’s surrounding environment.ExamplesSystem is just the coinFriction, weight, normal forceAre external forcesm1m21) System is m1, m2, the cord and pulleyweights, 2) N, F, and friction are externalTension forces are internal3) Choose m1and m2as separate systems.Tension becomes external to each systemCopyright R. Janow – Spring 2012Work done on an object by a constant force FOne dimensional world:startstopF∆∆∆∆xF and ∆∆∆∆x in same directionpositivex F W∆∆∆∆====startstopF∆∆∆∆xF and ∆∆∆∆x in opposite directionnegativex F- W∆∆∆∆====Two dimensional world:startstopFrxr∆∆∆∆θθθθ• Only the component of F along the displacement (along x) does work• The y-component of F does zero work (∆∆∆∆y = 0)• If several forces act, add up work contributions (+ or -) due to each.Example: a bead moving along x on a straight wire x )Fcos( W∆∆∆∆θθθθ====Units for Work and EnergyForce x Displacement = Work = [KE] = ½ mv2SI N x m = Joule = kg.m2/s2CGS Dyne x cm = Erg = gm.cm2/s2British lb x foot = foot-lb = slug ft2/s2Copyright R. Janow – Spring 2012Used to represent effect of force/displacement geometry on work done ) ABcos( BAoθθθθ≡≡≡≡rr- Vector times vector  scalar - Projection of A on B or B on A- Result = 0 if A & B are perpendicular- Result = |A|.|B| if A & B are parallel- CommutativeθθθθArBrComponent of B on AComponent of A on BDot Product Definition:Dot product (Scalar product, Inner product): AB BA oorrrr====- Distributive:C A B A ]C B A o o[ orrrrrrr++++====++++1kˆkˆ 1,jˆjˆ 1,iˆiˆ 0kˆiˆ 0, kˆjˆ 0, jˆiˆ======================== oooooo B A B A BA BAzzyyxxo++++++++====rrFor Cartesian Unit Vectors:For Vectors expanded in Cartesian Unit Vectors: kˆ A jˆ A iˆA Azyx++++++++====r kˆB jˆB iˆB Bzyx++++++++====r A A A A AAzyx2o222++++++++====≡≡≡≡rrCopyright R. Janow – Spring 2012Example: Dot Product using Cartesian Coordinates kˆ0 jˆ2 iˆ1 B ++++++++−−−−====r kˆ0 jˆ3 iˆ A ++++++++==== 2rVectors and are given in terms of unit vectors: Ar Bra) Calculate the scalar (dot) product : BAorr 0 2 3 (-1)2 B A B A BA BAzzyyxxo++++××××++++××××====++++++++====rrTo derive formula above use dot products of the unit vectors 4 BA o====rrb) Find the angle θθθθ between and :Ar Br ) ABcos( BAoθθθθ≡≡≡≡rr ABBA )cos(orr≡≡≡≡θθθθfrom the definitionFound above. Need magnitudes A and B BAorrUse Pythagorean Theorem or Dot products as follows:13 0 3 2 A A A A AA22zyx2o====++++++++====++++++++====≡≡≡≡222rr5 0 2 1)( B B B B BB22zyx2o====++++++++−−−−====++++++++====≡≡≡≡222rr 13 A ==== 5 B ====5134 ABBA )cos(o========θθθθrr 60.3 )134(cos o1-========θθθθ5Copyright R. Janow – Spring 2012Work Done by Constant Force, Continued rF )rcos(F W rov∆∆∆∆≡≡≡≡θθθθ∆∆∆∆====An object undergoes displacement in an arbitrarydirection while acted on by a constant single force .The work done ON the system by is: r r∆∆∆∆ F v F v• Only the force component parallelor antiparallel to the displacement does work• Force components perpendicular to thedisplacement are workless (e.g., centripetalforce)Note: If is small enough, is approximately constant r r∆∆∆∆F vdone work positive 90 0 orF ⇒⇒⇒⇒<<<<θθθθ⇒⇒⇒⇒>>>>∆∆∆∆rovdone work negative 90 0 orF ⇒⇒⇒⇒>>>>θθθθ⇒⇒⇒⇒<<<<∆∆∆∆rovdone work zero 90 0 orF ⇒⇒⇒⇒====θθθθ⇒⇒⇒⇒====∆∆∆∆rovKinetic energy of a particle mv K 221≡≡≡≡Copyright R. Janow – Spring 2012Work Done by a Constant Force6-1: The figure shows four examples of force F applied to an object. In all four cases, the force has the same magnitude and the displacement of the object is to the right and has the same magnitude. Rank the cases in order of the work done by the force on the object, from most positive to the most