Physics 151: Lecture 12

Physics 151: Lecture 12 • Topics : • (Text Ch. 7.1-7.4) • Work & Energy. • Work of a constant force. • Work of a non-constant force. • Work - Energy Theorem.

Forms of Energy • Kinetic: Energy of motion. • A car on the highway has kinetic energy. • We have to remove this energy to stop it. • The breaks of a car get HOT! • This is an example of turning one form of energy into another (thermal energy).

See text: 7-1 Energy Conservation • Energy cannot be destroyed or created. • Just changed from one form to another. • We say energy is conserved ! • True for any isolated system. • i.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. • The energy of the car “alone” is not conserved... • It is reduced by the braking. • Doing “work” on an otherwise isolated system will change it’s “energy”... Animation

See text: 7-1 Definition of Work: Ingredients: Force (F), displacement ( r) Work, W, of a constant force F acting through a displacement r is: W = F .r = F rcos  = Frr F r  Fr displacement “Dot Product”

F r  Fcos  Definition of Work... • Only the component of F along the displacement is doing work. • Example: Train on a track.

See text: 7.2 Review: Scalar Product ( or Dot Product) a Definition: a.b = ab cos  = a[b cos ] = aba = b[a cos ] = bab Some properties: a .b =b .a q(a .b) = (qb) .a = b .(qa)(q is a scalar) a .(b + c) = (a .b)+ (a .c)(c is a vector) The dot product of perpendicular vectors is 0 !! ba  b a  b ab

y j x i k z See text: 7.2 Review: Examples of dot products i .i = j .j = k .k = 1 i .j = j .k = k .i = 0 Suppose Then a = 1 i + 2 j + 3 k b = 4 i - 5 j + 6 k a.b = 1x4 + 2x(-5) + 3x6 = 12 a.a = 1x1 + 2x2 + 3x3 = 14 b.b = 4x4 + (-5)x(-5) + 6x6 = 77

a ay ax j i See text: 7.2 Review: Properties of dot products • Magnitude: a2 = |a|2 = a .a =(axi + ay j ) .(axi + ayj ) = ax2( i .i ) + ay2( j .j ) + 2axay( i .j ) = ax2 + ay2 • Pythagorian Theorem !!

Review: Properties of dot products • Components: a = ax i + ay j + az k = (ax , ay , az ) = (a. i , a. j , a. k ) • Derivatives: • Apply to velocity • So if v is constant (like for UCM):

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

F x See text: 7-1 Work: 1-D Example (constant force) • A force F= 10Npushes a box across a frictionless floor for a distance x = 5m. Work done byF on box : WF=F .x=Fx (since F is parallel to x) WF = (10 N)x(5m) = 50 N-m. See example 7-1: Pushing a trunk.

mks cgs other BTU = 1054 J calorie = 4.184 J foot-lb = 1.356 J eV = 1.6x10-19 J Dyne-cm (erg) = 10-7 J N-m (Joule) See text: 7-1 Units: Force x Distance = Work Newton x [M][L] / [T]2 Meter = Joule [L] [M][L]2 / [T]2

Text : 7.3 Work and Varying Forces • Consider a varying force, • DW = FxDx • As Dx  0, Dx  dx Area = FxDx Fx x Dx

Text : 7.3 Springs • A very common problem with a variable force is a spring. • In this spring, the force gets greater as the spring is further compressed. • Hook’s Law, FS = - kDx Dx is the amount the spring is stretched or compressed from it resting position. F Dx Animation

Lecture 12,ACT 2Hook’s Law • Remember Hook’s Law, Fx = -kDx What are the units for the constant k ? A) B) C) D)

Lecture 12,ACT 3Hook’s Law 8 cm 9 cm Whatis k for this spring ?? 0.2 kg A) 50 N/m B) 100 N/m C) 200 N/m D) 400 N/m

F(x) x1 x2 x Ws Ws -kx What is the Work done by the Spring... • The work done by the spring Wsduring a displacement from x1to x2 is the area under the F(x) vs x plot between x1and x2. kx1 kx2 Ws = - 1/2 [ ( kx2) (x2) - (kx1) (x1) ]

Work & Kinetic Energy: • A force F= 10Npushes a box across a frictionlessfloor for a distance x = 5m. The speed of the box is v1 before the push, and v2 after the push. v1 v2 F m i x

v1 v2 F x Work & Kinetic Energy... • Since the force F is constant, acceleration awill be constant. We have shown that for constant a: • S W = (S F).d = ma.d • For constant a, a =(v-v0)/t • also,d = vavt = 1/2 (v+v0)t m a i

v1 v2 F x Work & Kinetic Energy... • Altogether, • S W = (S F).d = ma.d =m [(v-v0)/t) .(1/2 (v+v0)t] • S W = 1/2 m ( v2 - v02 ) = (1/2 m v2 ) - (1/2 m v02 ) • Define Kinetic Energy K: K = 1/2mv2 • K2 - K1 = WF • WF = K (Work kinetic-energy theorem) m a i

See text: 7-4 Work Kinetic-Energy Theorem: NetWork done on object=change in kinetic energyof object • Is this applicable also for a variable force ? • YES

Lecture 12, ACT 4Kinetic Energy • 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)1/2 (c)1 (d) 2 (e) 4

to vo F m x V=0 t m ExampleWork Kinetic-Energy Theorem • How much will the spring compress to bring the object to a stop if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ? spring at an equilibrium position Dx = ( m vo2 / k ) 1/2 spring compressed

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) KA = 4 KB b) KA = 2 KB c) KA = KB d) KB = 2 KA e) KB = 4 KA

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) vA = 3/2 vB b) vA = (3/2 )1/2 vB c) vA = vB d) vB = (3/2 )1/2 vA e) vB = 3/2 vA

Recap of today’s lecture • Work & Energy. (Text: 7.1-4) • Discussion. • Definition. • Work of a constant and non-constant forces. • Work - Energy Theorem.

Physics 151: Lecture 12