Physics 121, Sections 9, 10, 11, and 12 Lecture 14 Today's Topics Homework 6 Chap6:#6,12,20,24,29,38,52,57,78,and83 Midterm 1: solutions Chapter 6: Work and Energy Review: Kinetic and Potential energy Non-conservative forces Generalized work-kinetic energy theorem Power Physics 121: Lecture 14, Pg 1
Physics 121: Lecture 14, Pg 1 Physics 121, Sections 9, 10, 11, and 12 Lecture 14 Today’s Topics: Homework 6: Chap. 6: # 6, 12, 20, 24, 29, 38, 52, 57, 78, and 83. Midterm 1: solutions Chapter 6: Work and Energy Review: Kinetic and Potential energy Non-conservative forces Generalized work-kinetic energy theorem Power
Definition of work Ingredients: Force(F), displacement (Ar) Work W. of a constant force F acting through a displacement A r is F W=F△rcos0 0/△r Physics 121: Lecture 14, Pg 2
Physics 121: Lecture 14, Pg 2 Definition of Work: Ingredients: Force ( F ), displacement ( r ) Work, W, of a constant force F acting through a displacement r is: W = F r cos F r Fr
Work Kinetic-Energy Theorem INet Work done on object] Change in kinetic energy of object W=△K K-K mv mV Physics 121: Lecture 14, Pg 3
Physics 121: Lecture 14, Pg 3 Work Kinetic-Energy Theorem: {Net Work done on object} = {change in kinetic energy of object} Wnet = K = K2 − K1 2 1 2 2 mv 2 1 mv 2 1 = −
Potential Energy Recap For any conservative force we can define a potential energy function U such that AU=U-U=-W The potential energy function U is always defined only up to an additive constant You can choose the location where u=o to be anywhere convenient Physics 121: Lecture 14, Pg 4
Physics 121: Lecture 14, Pg 4 Potential Energy Recap: For any conservative force we can define a potential energy function U such that: The potential energy function U is always defined only up to an additive constant. You can choose the location where U = 0 to be anywhere convenient. U = U2 - U1 = - W
Conservative Forces Potential Energies (stuff you should know): Force Work Change in P.E P.E. function F W(1-2) △U=U2-U Fg=-mg j -mg(y2-yi) mg) mgy +C GMm GMm GMm GMm IR2 R 尺,R R Fs=-kX KIx Physics 121: Lecture 14, Pg 5
Physics 121: Lecture 14, Pg 5 Conservative Forces & Potential Energies (stuff you should know): Force F Work W(1-2) Change in P.E U = U2 - U1 P.E. function U Fg = -mg j Fg = r Fs = -kx ^ ^ − 2 R1 1 R 1 GMm − − 1 2 2 2 1 2 k x x -mg(y2 -y1 ) mg(y2 -y1 ) − − GMm R R 1 1 2 1 1 2 2 2 1 2 k x − x − GMm R 2 mgy + C C R GMm − + kx +C 2 2 1
Conservation of Energy If only conservative forces are present, the total energy (sum of potential and kinetic energies ) of a system is conserved E=K+U △E=△K+△U =W+△U using△K=W W+(-W=0 using△U=-W E=K+ U is constant∥! Active Figure Both K and U can change but E=K+Uremains constant Physics 121: Lecture 14, Pg 6
Physics 121: Lecture 14, Pg 6 Conservation of Energy If only conservative forces are present, the total energy (sum of potential and kinetic energies) of a system is conserved. E = K + U E = K + U = W + U = W + (-W) = 0 using K = W using U = -W Both K and U can change, but E = K + U remains constant. E = K + U is constant !!! Active Figure
Problem: Hotwheel a toy car slides on the frictionless track shown below It starts at rest, drops a distance d, moves horizontally at speed V,, rises a distance h, and ends up moving horizontally with speed v, Find v, and v2 h Physics 121: Lecture 14, Pg 7
Physics 121: Lecture 14, Pg 7 Problem: Hotwheel. A toy car slides on the frictionless track shown below. It starts at rest, drops a distance d, moves horizontally at speed v1 , rises a distance h, and ends up moving horizontally with speed v2 . Find v1 and v2 . h d v1 v2
Problem: Hotwheel Energy is conserved, So AE=0 △KE=-△PE Moving down a distance d,△PE=mg,△KE=1mv2 Solving for the speed 2gd h Physics 121: Lecture 14, Pg 8
Physics 121: Lecture 14, Pg 8 Problem: Hotwheel... Energy is conserved, so E = 0 KE = - PE Moving down a distance d, PE = -mgd, KE = 1 /2mv1 2 Solving for the speed: h d v1 v1 = 2gd
Problem: Hotwheel At the end we are a distance d-h below our starting point △PE=-mg(d-h),△KE=2mv2 Solving for the speed 2g(d-h d-h h Physics 121: Lecture 14, Pg 9
Physics 121: Lecture 14, Pg 9 Problem: Hotwheel... At the end, we are a distance d-h below our starting point. PE = -mg(d-h), KE = 1 /2mv2 2 Solving for the speed: h d v2 v g(d h) 2 = 2 − d-h
Lecture 14, Example Skateboard What speed will skateboarder reach at bottom of the hill? Initial: K,=0 U1=mgR Final: K = 1/2 mv2 U=0 m= 25 kg Conservation of Total Energy J1=K2+U2 R=3 m 0+mgR=1/2mM2+0 (2gR)12 V~(2x10m/s2X3m)12 v-8 m/s(-16mph) Does not depend on the mass Physics 121: Lecture 14, Pg 10
Physics 121: Lecture 14, Pg 10 What speed will skateboarder reach at bottom of the hill ? R=3 m .. m = 25 kg Initial: K1 = 0 U1 = mgR Final: K2 = 1/2 mv2 U2 = 0 Conservation of Total Energy : Lecture 14, Example Skateboard K1 + U1 = K2 + U2 0 + mgR = 1/2mv2 + 0 v = (2gR)1/2 v ~ (2 x10m/s2 x 3m)1/2 v ~ 8 m/s (~16mph) ! Does NOT depend on the mass !