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Unit 07 “Work, Power, Energy and Energy Conservation” The Conservation of Mechanical Energy Problem Solving.

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Presentation on theme: "Unit 07 “Work, Power, Energy and Energy Conservation” The Conservation of Mechanical Energy Problem Solving."— Presentation transcript:

1 Unit 07 “Work, Power, Energy and Energy Conservation” The Conservation of Mechanical Energy Problem Solving

2 Solving Problems Using the Conservation of Mechanical Energy ME i = ME f The total mechanical energy for a system must stay the same (unless there is friction) KE i + PE gi KE f + PE gf ½mv i 2 mgh i ½mv f 2 mgh f = = + +

3 ME i = ME f Solving Problems Using the Conservation of Mechanical Energy The total mechanical energy for a system must stay the same (unless there is friction) KE f + PE gf KE i + PE gi ½mv i 2 mgh i ½mv f 2 mgh f

4 Before you begin the “Problem Solving Steps”, you should: Draw a picture and identify the initial and final points. Then identify what types of energy the object has at the initial and final points. A car starts at rest on a hill and then begins to roll down…

5 A ball is thrown and moves upward to its final height … Before you begin the “Problem Solving Steps”, you should: Draw a picture and identify the initial and final points. Then identify what types of energy the object has at the initial and final points.

6 A dog starts from rest at the top of a hill and begins to slide down, at the middle of the hill… Before you begin the “Problem Solving Steps”, you should: Draw a picture and identify the initial and final points. Then identify what types of energy the object has at the initial and final points.

7 Medium Problem Solving A 2000kg car is at rest on the top of 100m high hill, how fast is it going at the bottom?

8 Medium Problem Solving 1. A 2000kg car is at rest on the top of 100m high hill, how fast is it going at the bottom? ME i = ME f KE i + PE gi KE f + PE gf mgh i ½mv f 2 = = PE gi = KE f (2000kg)(9.8m/s 2 )(100m)½(2000kg)v f 2 = 1960000J = 1000kgv f 2 1960m 2 /s 2 = vf2vf2 44.3 m/s = vfvf m= 2000kg h i =100m V f =? (V i =0m/s) (h f =0m)

9 2. A 5kg ball is thrown upward with a velocity of 4m/s. How high does the ball go?

10 ME i = ME f KE i + PE gi ½mv i 2 mgh f = = KE i = PE gf = 40 J = ½ (5kg)(4m/s) 2 0.816m = hfhf KE f + PE gf (5kg)(9.8m/s 2 )(h f ) (49 kgm/s 2 )(h f ) m= 5kg V i =4m/s h f =? (h i =0m) (V f =0m/s)

11 Hard A 2000kg car starts from rest at the top of a hill, it reaches a speed of 34.3m/s at a point 40m above the ground. How is the hill where the car starts?

12 Hard (example) ME i = ME f KE i + PE gi KE f + PE gf mgh i ½mv f 2 + = = PE gi = (2000kg)(9.8m/s 2 )(h i ) ½(20000kg)(34.3m/s) 2 = (19600kgm/s 2 )(h i ) = 1176490 J + 19600kgm/s 2 (h i ) = 1960490 J Initial (top) m= 2000kg h i =? KE f + PE gf mgh f + (2000kg)(9.8m/s 2 )(40m) 784000 J hihi = 100 m Final (middle) m=2000kg V f =34.3m/s h f =40m

13 1 st Find the kinetic energy at the initial position 2 nd Find the total mechanical energy at the initial position. 3 rd What is the total mechanical energy at the final position? 4 th What is the gravitational potential energy at the final position? 5 th What is the height of the ball at the final position? What is the rock’s total mechanical energy at any point?

14 KE i = ½ mv i 2 KE i = ½ (5kg)(4m/s) 2 KE i = 40 J ME i = PE gi +KE i ME i = 0 J + 40 J ME i = 40 J ME i = ME f 40 J = ME f 1 st Find the kinetic energy at the initial position 2 nd Find the total mechanical energy at the initial position. 3 rd What is the total mechanical energy at the final position? 4 th What is the gravitational potential energy at the final position? ME f = PE gf +KE f 40 J = PE gf + 0J 40 J = PE gf 5 th What is the height of the ball at the final position? PE gf = mgh f 40 J = (5kg)(9.8m/s 2 )h f 40 J = (49kgm/s 2 )h f 0.816m= h f What is the rock’s total mechanical energy at any point? 40 J

15 Quiz Review How high does a 6kg ball go if it s thrown from the group upwards with an initial velocity of 10m/s?


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