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The Law of Conservation of Energy

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Follow Me Suppose a 0.500kg rock falls from a height of 78.4m and that its Eg at the bottom is zero (h=0). 1.Its speed is: 2. Distance fallen is:

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At the Top 3. Height is: 4. Its gravitational potential energy at the top will be: 5. Its kinetic energy at the top will be 0.

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At the Bottom 6. At the bottom, Eg=0 (h=0) and kinetic energy is:

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In The Middle 7. In the middle: Eg+Ek=192+192=384J

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Law of Conservation of Energy Energy can not be created or destroyed, only changed from one form to another In any transfer or transformation of energy, the total amount of energy remains constant. Energy is not necessarily changed into the form you want (ex. Thermal energy produced by friction)

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Conservation of Energy Video Wrecking Ball with Teacher (involves conservation of energy) http://www.youtube.com/watch?v=mhIOylZMg6Q

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Example As the water in a river approaches a 12.3 m vertical drop, its average speed is 6.7 m/s. For each kilogram of water in the river, determine the following: a) the kinetic E at the top of the waterfall b) the gravitational potential energy at the top of the falls relative to the bottom c) the total mechanical E at the bottom of the falls, not considering friction (use the law of conservation of E) d) the speed at the bottom of the falls (use the law of conservation of E) (v=16.9m/s)

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Challenge Take a piece of gum, candy, anything with nutritional information Using the energy content, how high could you throw a 10kg metal ball if your body converted 100% of this energy to the ball? NOTE: You will need to use a conversion website to choose the appropriate conversion factor (be careful to note if calories is capitalized or not)

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Practice Problems 1.The toy car below continues to move and rolls off the table. Find the cars speed just before it hits the floor. [6.49m/s] 2. The speed of an acrobat swinging on a trapeze is 5.64 m/s at the lowest point of her motion. Assume her mass is 53.7 kg. (a) How high above the lowest point can she swing? [1.62m] (b) Do you need to know her mass to answer part (a)? Explain.

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Practice Problems 3. (a) When you flip a penny (2.35 g), it leaves your hand and moves upward at 2.85 m/s. Use energy to find how high the penny goes above your hand before stopping. [0.414m] (b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor. [5.73m/s] (c) Explain your choice of reference level for parts (a) and (b). (d) Choose a different reference level and repeat part (b). 4. A roller-coaster train and its passengers have a combined mass of 1250 kg. The train comes over the top of the first hill, 53.2 m above the ground, with a speed of 1.17 m/s. (a) The train goes down the first hill and through a loop. Ignoring friction, calculate the speed of the train at the top of the loop, 21.3 m above the ground. [25.0m/s] (b) Before applying its brakes at the end of the ride, the train moves along a level stretch of track 2.71 m above the ground. The trains speed is 24.3 m/s. How much mechanical energy has been lost during the ride? Where did this energy go? [2.51 x 10 5 J]

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Research Applications 1.Research into a machine or device that uses the law of conservation of mechanical energy to perform its primary function. Explain how the law is applied using specific, relevant terminology from this unit. 2.Research into another machine or device that uses the law of conservation of energy to perform its primary function, but not transforming mechanical energy. Explain how the law is applied using specific, relevant terminology from this unit.

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Extra Help Everything you need to know from this lesson: http://www.youtube.com/watch?v=iYEWIuQBVyg

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Efficiency

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Sample Problem A wheel and axle is used to lift a 10 kg load to a vertical height of 30 cm. The force used to lift this load is 33 N and the force acts through a distance of 1.2 m. a. Determine the work done on the load b. Calculate the efficiency of this system.

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Sample Problem Solution a. E in = W = Fd = (33 N)(1.2 m) = 40 J b. E out = the potential E created by raising the load E out = E p = mgh = (10 kg)(9.8 N/kg)(0.30 m) = 29 J efficiency = E out x 100% E in = 29 J x 100% 40 J = 73 %