Physics 151: Lecture 12, Pg 1 Physics 151: Lecture 12 l Topics : ç( Text Ch ) »Work & Energy. »Work of a constant force. »Work of a non-constant force. »Work - Energy Theorem.

Physics 151: Lecture 12, Pg 2 Forms of Energy l Kinetic l Kinetic: Energy of motion. çA car on the highway has kinetic energy. çWe have to remove this energy to stop it. çThe breaks of a car get HOT ! çThis is an example of turning one form of energy into another (thermal energy).

Physics 151: Lecture 12, Pg 3 Energy Conservation l Energy cannot be destroyed or created. çJust changed from one form to another. energy is conserved l We say energy is conserved ! çTrue for any isolated system. çi.e when we put on the brakes, the kinetic energy of the car is turned into heat using friction in the brakes. The total energy of the “car-breaks-road-atmosphere” system is the same. çThe energy of the car “alone” is not conserved... »It is reduced by the braking. work energy l Doing “work” on an otherwise isolated system will change it’s “energy”... See text: 7-1 Animation

Physics 151: Lecture 12, Pg 4 Definition of Work: Ingredients: Fr Ingredients: Force ( F ), displacement (  r ) F Work, W, of a constant force F r acting through a displacement  r is: F rrr W = F.  r = F  r cos  = F r  r  F r rr r displacement FrFr “Dot Product” See text: 7-1

Physics 151: Lecture 12, Pg 5 Definition of Work... l Only the component of F along the displacement is doing work. çExample: Train on a track.F r rr r  F cos 

Physics 151: Lecture 12, Pg 6 Review: Scalar Product ( or Dot Product) Definition: ab a. b= ab cos  = a[b cos  ] = ab a = b[a cos  ] = ba b Some properties: a bb a a. b= b. a a bb a b a q(a. b) = (qb). a = b. (qa) (q is a scalar) a b ca b a c c a. (b + c) = (a. b) + (a. c) (c is a vector) The dot product of perpendicular vectors is 0 !! See text: 7.2  a abab b  a b baba

Physics 151: Lecture 12, Pg 7 Review: Examples of dot products SupposeThen a i j k a = 1 i + 2 j + 3 k b i j k b = 4 i - 5 j + 6 k ab a. b = 1x4 + 2x(-5) + 3x6 = 12 aa a. a = 1x1 + 2x2 + 3x3 = 14 bb b. b = 4x4 + (-5)x(-5) + 6x6 = 77 i i j j k k i. i = j. j = k. k = 1 i j j k k i i. j = j. k = k. i = 0 See text: 7.2 x y z i j k

Physics 151: Lecture 12, Pg 8 Review: Properties of dot products l Magnitude: a 2 = |a| 2 = a. a i j i j = (a x i + a y j ). (a x i + a y j ) i i j j i j = a x 2 ( i. i ) + a y 2 ( j. j ) + 2a x a y ( i. j ) = a x 2 + a y 2 çPythagorian Theorem !! a axax ayay i j See text: 7.2

Physics 151: Lecture 12, Pg 9 Review: Properties of dot products l Components: a i j ka ia ja k a = a x i + a y j + a z k = (a x, a y, a z ) = (a. i, a. j, a. k ) l Derivatives: çApply to velocity çSo if v is constant (like for UCM):

Physics 151: Lecture 12, Pg 10 Lecture 12, ACT 1 Work A box is pulled up a rough (  > 0) incline by a rope-pulley- weight arrangement as shown below. çHow many forces are doing work on the box ? (a) (a) 2 (b) (b) 3 (c) (c) 4 Is the work done by F positive or negative?

Physics 151: Lecture 12, Pg 11 Work: 1-D Example (constant force) F x A force F = 10N pushes a box across a frictionless floor for a distance  x = 5m. xxxxF byF on Work done by F on box : F xFx W F = F.  x = F  x (since F is parallel to  x) W F = (10 N)x(5m) = 50 N-m. See text: 7-1 See example 7-1: Pushing a trunk.

Physics 151: Lecture 12, Pg 12 Units: N-m (Joule) Dyne-cm (erg) = J BTU= 1054 J calorie= J foot-lb= J eV= 1.6x J cgsothermks Force x Distance = Work Newton x [M][L] / [T] 2 Meter = Joule [L] [M][L] 2 / [T] 2 See text: 7-1

Physics 151: Lecture 12, Pg 13 Work and Varying Forces l Consider a varying force,  W = F x  x As  x  0,  x  dx Text : 7.3 FxFx x xx Area = F x  x

Physics 151: Lecture 12, Pg 14 Springs l A very common problem with a variable force is a spring. l In this spring, the force gets greater as the spring is further compressed. l Hook’s Law, F S = - k  x  x is the amount the spring is stretched or compressed from it resting position. Text : 7.3 F xx Animation

Physics 151: Lecture 12, Pg 15 Lecture 12, ACT 2 Hook’s Law l Remember Hook’s Law, F x = -k  x What are the units for the constant k ? A)B)C)D)

Physics 151: Lecture 12, Pg 16 Lecture 12, ACT 3 Hook’s Law 0.2 kg 9 cm 8 cm What is k for this spring ?? A) 50 N/mB) 100 N/mC) 200 N/m D) 400 N/m

Physics 151: Lecture 12, Pg 17 What is the Work done by the Spring... x2x2 x1x1 F(x) x l The work done by the spring W s during a displacement from x 1 to x 2 is the area under the F(x) vs x plot between x 1 and x 2. WsWs kx 1 kx 2 -kx WsWs W s = - 1/2 [ ( kx 2 ) (x 2 ) - (kx 1 ) (x 1 ) ]

Physics 151: Lecture 12, Pg 18 Work & Kinetic Energy: F x A force F = 10N pushes a box across a frictionless floor for a distance  x = 5m. The speed of the box is v 1 before the push, and v 2 after the push. xxxx F v1v1 v2v2 i m

Physics 151: Lecture 12, Pg 19 Work & Kinetic Energy... Fa l Since the force F is constant, acceleration a will be constant. We have shown that for constant a:   W = (  F). d = ma. d çFor constant a, a = (v-v 0 )/t çalso, d = v av t = 1/2 (v+v 0 )t xxxx F v1v1 v2v2a i m

Physics 151: Lecture 12, Pg 20 Work & Kinetic Energy... l Altogether,   W = (  F). d = ma. d = m [(v-v 0 )/t). (1/2 (v+v 0 )t]   W = 1/2 m ( v 2 - v 0 2 ) = (1/2 m v 2 ) - (1/2 m v 0 2 ) l Define Kinetic Energy K:K = 1 / 2 mv 2 çK 2 - K 1 = W F (Work kinetic-energy theorem)  W F =  K (Work kinetic-energy theorem) xxxx F v1v1 v2v2a i m

Physics 151: Lecture 12, Pg 21 Work Kinetic-Energy Theorem: NetWorkchangekinetic energy Net Work done on object = change in kinetic energy of object See text: 7-4 l Is this applicable also for a variable force ? l YES

Physics 151: Lecture 12, Pg 22 Lecture 12, ACT 4 Kinetic Energy l To practice your pitching you use two baseballs. The first time you throw a slow curve and clock the speed at 50 mph (~25 m/s). The second time you go with high heat and the radar gun clocks the pitch at 100 mph. What is the ratio of the kinetic energy of the fast ball versus the curve ball ? (a) (b) (c) (a) 1/4 (b) 1/2 (c) 1 (d) 2 (e) 4

Physics 151: Lecture 12, Pg 23 xxxx vovo m toto F Example Work Kinetic-Energy Theorem  x = ( m v o 2 / k ) 1/2  How much will the spring compress to bring the object to a stop if the object is moving initially at a constant velocity (v o ) on frictionless surface as shown below ? spring compressed spring at an equilibrium position V=0 t m

Physics 151: Lecture 12, Pg 24 Act 4 Two clowns are launched from the same spring-loaded circus cannon with the spring compressed the same distance each time. Clown A has a 40-kg mass; clown B a 60-kg mass. The relation between their kinetic energies at the instant of launch is : a)K A = 4 K B b) K A = 2 K B c) K A = K B d) K B = 2 K A e) K B = 4 K A

Physics 151: Lecture 12, Pg 25 Act 4.b Two clowns are launched from the same spring-loaded circus cannon with the spring compressed the same distance each time. Clown A has a 40-kg mass; clown B a 60-kg mass. The relation between their speeds at the instant of launch is: a)v A = 3/2 v B b) v A = (3/2 ) 1/2 v B c) v A = v B d) v B = (3/2 ) 1/2 v A e) v B = 3/2 v A

Physics 151: Lecture 12, Pg 26 Recap of today’s lecture l Work & Energy. (Text: 7.1-4) çDiscussion. çDefinition. l Work of a constant and non-constant forces. l Work - Energy Theorem.