1. 10/4 Potential Energy Intro  Text: Chapter 6 Energy  HW 10/4 “Potential Energy with Friction” due Wednesday 10/9 (to be posted soon)  Potential Energy and Projectile Motion  Discuss “Block up a Ramp”  Discuss “2nd Law vs. Energy”  Exam 2 Thursday, 10/17 5-7 Wit 116 6-8 Wit 114 (only if needed)

2. Example: Energy in 2D 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,y = 2m/s “Work” KE i KE f 6J 90J 96J

3. 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)

4. Energy in 2D  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) What if F net and  x point in different directions?

5. Example: Energy in 2D Energy 3.0m v f,y = 8m/s 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,y = 2m/s “Work” KE i KE f 6J 90J 96J Let’s just add a horizontal component to the velocity. v i,x = 6m/s v f = 10m/s The y-component solution does not change.

6. Example: Energy in 2D Energy 3.0m v f,y = 8m/s Watch what happens when we ignore the vectors! KE i = 1 / 2 mv i 2 = m = 3kg Work = mg  y = KE f = = 1 / 2 mv f 2 v f = W E,B =F net = mg v i,y = 2m/s “Work” KE i KE f 60J 90J 150J Let’s just add a horizontal component to the velocity. v i,x = 6m/s v f = 10m/s The y-component solution does not change. v i =  40 m/s 60J 90J ??? 150J 10m/s !!!!!

7. Potential Energy Energy 3.0m v f,y = 8m/s Watch what happens when we ignore the vectors! KE i = 1 / 2 mv i 2 = m = 3kg Work = mg  y = KE f = = 1 / 2 mv f 2 v f = W E,B =F net = mg v i,y = 2m/s “Work” KE i KE f 60J 90J 96J Since this worked, we choose to define “Potential Energy” as.. v i,x = 6m/s v f = 10m/s v i =  40 m/s 60J 90J ??? 150J 10m/s !!!!! PE g = mgy  PE g = mg  y

8. Watch Out for +- Signs!!!  PE in the previous case is negative, corresponding to a drop in PE and a rise in KE. So- a drop in PE will mean a rise in KE and a rise in PE will mean a drop in KE.  KE = -  PE Here the minus signs mean gain or loss, not direction.