10/3 Energy Intro  Text: Chapter 6 Energy  HW 10/3 “2nd Law vs. Energy” due Monday 10/7  Tomorrow, Potential Energy and Projectile Motion  Exam 2 Thursday, 10/ Wit Wit 114 (only if needed)  Use 5kg and 30N in “Block on a Ramp” but try it with all combinations of numbers!

Energy Introduction The equation I never let you use: v i 2 + 2a  x = v f 2 Multiply by the mass of the object, m: mv i 2 + 2ma  x = mv f 2 Divide by 2: 1 / 2 mv i 2 + ma  x = 1 / 2 mv f 2 Note that ma = F net :(2nd Law) 1 / 2 mv i 2 + F net  x = 1 / 2 mv f 2 Defines Kinetic Energy!

Kinetic Energy and “Work” 1 / 2 mv i 2 + F net  x = 1 / 2 mv f 2 Defines Kinetic Energy! KE i + “Work” = KE f Shorthand for “Work Energy Relation” This is not a vector equation, it’s a “Bucket” equation! Think of “Work” as the “change in Energy” or the amount of fluid I pour into (or out of) a Bucket Think of Energy as a “fluid” “Work” KE i KE f

Example: 2nd law vs. Energy 2nd Law 3.0m v i = 2m/s v f = ? Wish for an easier way!!! m = 3kg W E,B =F net = mg Let’s see…. find acceleration Find the final velocity for a ball moving up at 2m/s when it is 3m below its starting point. find time to top find  y to top find total  y find v f

Example: 2nd law vs. Energy Energy 3.0m v f = ? Let’s see… Find KE i ( 1 / 2 mv i 2 ) Find “Work” (F net  y) Add to Ke i to get KE f Find v f from KE f KE i = 1 / 2 mv i 2 = 6 Joules m = 3kg Work = mg  y = 90 Joules KE f = 96J = 1 / 2 mv f 2 v f = 8.0m/s Oh Baby! W E,B =F net = mg v i = 2m/s “Work” KE i KE f 6J 90J 96J Benefits: Don’t need highest point! (  y = 3.0m) No acceleration calculation! No vectors!

Energy  Always draw “Buckets”  For “Work” the net force must be constant in F net  x   x is the “Displacement,” as usual  Always consider Energy as an alternative to the 2nd law  Energy is a “Scalar” not a “Vector” (See Ch. 1) For Friday, think about this example in the case where there is a horizontal component to the motion. (projectile motion)