Physics 1501: Lecture 11, Pg 1 Physics 1501: Lecture 11 l Announcements çHW 04 will be due the following Friday (same week). çMidterm 1: Monday Oct. 3 çWednesday »Replacement (Sandipan Banerjee) »Office hours Friday instead of Wednesday l Topics çWork & Energy çScalar Product

Physics 1501: Lecture 11, Pg 2 Chap.6: Work & Energy l One of the most important concepts in physics. çAlternative approach to mechanics. l Many applications beyond mechanics. çThermodynamics (movement of heat). çQuantum mechanics... l Very useful tools. çYou will learn new (sometimes much easier) ways to solve problems.

Physics 1501: Lecture 11, Pg 3 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”

Physics 1501: Lecture 11, Pg 4 Review: Scalar Product ( or Dot Product) Definition: a b a · b= ab cos  = a[b cos  ] = ab a = b[a cos  ] = ba b Some properties: a bb a a ·b= b ·a a bba 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 !!  a abab b  a b baba

Physics 1501: Lecture 11, Pg 5 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 x y z i j k

Physics 1501: Lecture 11, Pg 6 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

Physics 1501: Lecture 11, Pg 7 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 1501: Lecture 11, Pg 8 Back to the definition of Work: F Work, W, of a force F acting r through a displacement  r is: F r W = F ·  r F r rr r

Physics 1501: Lecture 11, Pg 9 Lecture 11, 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

Physics 1501: Lecture 11, Pg 10 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 F xFx Work done by F on box : W F = F ·  x = F  x (since F is parallel to  x) W F = (10 N)x(5m) = 50 N-m.

Physics 1501: Lecture 11, Pg 11 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

Physics 1501: Lecture 11, Pg 12 Work and Varying Forces l Consider a varying force,  W = F x  x As  x  0,  x  dx FxFx x xx Area = F x  x

Physics 1501: Lecture 11, Pg 13 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. F xx Active Figure

Physics 1501: Lecture 11, Pg 14 Lecture 11, 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 1501: Lecture 11, Pg 15 Lecture 11, 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 1501: Lecture 11, Pg 16 What is the Work done by the Spring... 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. W s = - 1/2 [ ( kx 2 ) (x 2 ) - (kx 1 ) (x 1 ) ] x2x2 x1x1 F(x) x WsWs kx 1 kx 2 -kx WsWs Active Figure

Physics 1501: Lecture 11, Pg 17 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 1501: Lecture 11, Pg 18 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 = vt = (1/2) (v+v 0 )t xxxx F v1v1 v2v2a i m

Physics 1501: Lecture 11, Pg 19 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 ) 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 1501: Lecture 11, Pg 20 Work Kinetic-Energy Theorem: NetWork { Net Work done on object } = changekinetic energy { change in kinetic energy of object } l We’ll prove this for a variable force later.

Physics 1501: Lecture 11, Pg 21 xxxx vovo m toto F Example Work Kinetic-Energy Theorem  K m = 1/2 m v 2 - 1/2 m v o 2 = 0 - 1/2 m v o 2  K m = W spring W spring = -1/2 k  x 2 - x o 2 ) = - 1/2 k  x 2 (assuming x 0 =0) so : 1/2 m v o 2 = 1/2 k  x 2 and :  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 1501: Lecture 11, Pg 22 Lecture 11, 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) 1/4 (b) (c) (a) 1/4 (b) 1/2 (c) 1 (d) 2 (e) 4

Physics 1501: Lecture 11, Pg 23 Lecture 11, ACT 5 Kinetic Energy l To practice your pitching you use two baseballs. The first time you use a softball: you throw a slow curve and clock the speed at 50 mph (~25 m/s). The second time you use a baseball (about half the mass of a softball) and go with high heat: 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 1501: Lecture 11, Pg 24 Lecture 11, ACT 6 Work & Energy Two blocks having mass m 1 and m 2 where m 1 > m 2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e.  > 0) which slows them down to a stop. Which one will go farther before stopping ? (a) (b) (c) (a) m 1 (b) m 2 (c) they will go the same distance m1m1 m2m2

Physics 1501: Lecture 11, Pg 25 Lecture 11, ACT 7 Work & Energy l You like to drive home fast, slam on your brakes at the bottom of the driveway, and screech to a stop laying rubber all the way. It’s particularly fun when your mother is in the car with you. You practice this trick driving at 20 mph and with some groceries in your car with the same mass as your mama. You find that you only travel half way up the driveway. Thus when your mom joins you in the car, you try it driving twice as fast. How far will you go this time ? (a)The same distance. Not so exciting. (b)  2 times as far (only 7/10 of the way up the driveway) (c) twice as far, right to the door. Whoopee! (d) four times as far. Crashes into house. Sorry Ma.