1. More Fun Energy Examples 1. Santa Claus is at the top of a chimney 8.5 m from the floor below. He slips and falls all the way down (ouch!). With what speed does he hit the floor? We will use Conservation of Energy Santa has potential energy at the top. How much PE does he have? PE = m g hm = ? Assume m = 150 kg Then PE = m g h= ( 150 kg )( 9.8 m/s 2 )( 8.5 m ) PE = 12 495 J

2. 1. Santa Claus is at the top of a chimney 8.5 m from the floor below. He slips and falls all the way down. With what speed does he hit the floor? PE top = 12 495 J All PE at the top is transformed into KE at the bottom so KE bottom = 12 495 J Then we can find speed from v = 2 KE m 2 ( 12 495 J ) 150 kg == v = 13 m/s

3. 1. Santa Claus is at the top of a chimney 8.5 m from the floor below. He slips and falls all the way down. With what speed does he hit the floor? Would we get a different answer if Santa trimmed down to 100 kg? PE = m g h= ( 100 kg )( 9.8 m/s 2 )( 8.5 m ) PE top = 8330 J KE bottom = PE top = 8330 J v = 2 KE m 2 ( 8330 J ) 100 kg == v = 13 m/s same result

4. 1. Santa Claus is at the top of a chimney 8.5 m from the floor below. He slips and falls all the way down. With what speed does he hit the floor? Conservation of Energy problems are mass-independent KE bottom = PE top For objects that fall from rest, ½ m v 2 = m g h ½ v 2 = g h v 2 = 2 g h v = 2 g h= 2 ( 9.8 m/s 2 )( 8.5 m ) = v = 13 m/s Speed of object dropped from height h 22 for this example

5. 2. A box is sliding on a frictionless surface at 3.6 m/s. It slides onto a frictionless ramp. How far above the floor does it rise? h = ? v = 3.6 m/s Use Conservation of Energy KE bottom = PE top ½ m v 2 = m g h ½ v 2 = g h v2v2 2 g = h gg == h = 0.66122 m ( 3.6 m/s ) 2 2 ( 9.8 m/s 2 ) h = 0.66 m

6. 3. The box in #2 slides back down the ramp and onto a surface that has a coefficient of friction of 0.25. How far does the box slide before coming to rest? 0.66 m μ = 0.25 d = ? h = 0.66122 m PE at top of ramp = m g h This PE is transformed into KE at the bottom of the ramp KE bottom = PE top = m g h Friction stops the box, taking away its KE ; friction does work on the box How much work does friction do?The same amount as its energy, m g h so Work Done by Friction = m g h

7. 3. The box in #2 slides back down the ramp and onto a surface that has a coefficient of friction of 0.25. How far does the box slide before coming to rest? 0.66 m μ = 0.25 d = ? h = 0.66122 m Work Done by Friction = m g h F f d = m g h μF N d = m g h μ mg d = m g h μ Wt. d = m g h μ d = h μμ d = μ h 0.25 ( 0.66122 m ) = d = 2.6 m