lecture15

lecture15 - Example A 55-kg skateboarder starts out with a...

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Unformatted text preview: Example : A 55-kg skateboarder starts out with a speed of 1.80 m/s. He does +80.0 J of work on himself by pushing with his feet against the ground. In addition, friction does - 265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is 6.00 m/s. (a) Calculate the change in the gravitational potential energy ( Δ PE = PE f – PE o ). (b) How much has the vertical height of the skateboarder changed? (c) Is he above or below the starting point? (a) o f E E- = NC W o o f f PE KE PE KE-- + = o f 2 2 1 2 2 1 PE PE-- + = o f mv mv PE Δ )- ( W PE 2 2 2 1 NC f o v v m + = Δ ) 6 8 . 1 )( 55 ( 265- 80 2 2 2 1- + = PE Δ J 1086- = (b) 1086 ) ( PE- =- = Δ o f h h mg mg h h o f 1086 ) (- =- ⇒ m 1 . 2 ) 8 . 9 )( 55 ( 1086- =- = (c) Since , , so he is below where he started . <- o f h h f o h h > 6.7 Power We’ve discussed what work is, but what about the time it takes to do the work??? Let’s say that both of us have 50 identical boxes to lift up on to a 1-m high table. We each have the same amount of work to do. But what if I lift my 50 boxes in 10 minutes, and you lift yours in 20 minutes. I’ve done the same amount of work as you, but I’ve done it in half the time. There must be a way of quantifying this....
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This note was uploaded on 08/04/2009 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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lecture15 - Example A 55-kg skateboarder starts out with a...

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