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The power of Niagra Falls

The power of Niagra Falls. Height: 167 ft Flow: 600,000 U.S. gallons per second. The power of Einstein. Kinetic energy: E = ½ mV 2 Energy of matter: E = mc 2. Potential Energy and Conservation of Mechanical Energy. CONSERVATIVE FORCES are forces that….. Produce only mechanical motion

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The power of Niagra Falls

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  1. The power of Niagra Falls Height: 167 ft Flow: 600,000 U.S. gallons per second

  2. The power of Einstein Kinetic energy: E = ½ mV2 Energy of matter: E = mc2

  3. Potential Energy and Conservation of Mechanical Energy • CONSERVATIVE FORCES are forces that….. • Produce only mechanical motion • Store energy in mechanical motion • Examples are gravity and spring forces. Chapter 8-1 / 8-3 • NON-CONSERVATIVE FORCES are forces that…. • Create energy in the form of heat, sound, or other non-mechanical process • Cause a transfer of energy from one system to another • Example is friction.

  4. Work of a conservative force can be positive or negative. W = F * D Dx Dx F F Work of gravity is NEGATIVE. Work of gravity is POSITIVE.

  5. Work done by a conservative force around a closed path is zero. This leads to an important conclusion…..

  6. A A B B The work done by a conservative force is independent of the path, and depends only on the starting and ending points. Pick any starting and ending points. Closed path, W=0. W2 W1 W1 = WAB W2 = WBA W1 + W2 = 0 W3 W1 W1 + W3 = 0 So, all paths from B to A take the same amount of work.

  7. Conservative forces produce “Potential Energy” W = - DU = -(Ufinal - Uinitial) W = FDx = -MgH Ufinal so MgH = (Ufinal - Uinitial) F Dx = H Mgyfi - Mgyin = (Ufinal - Uinitial) Uinitial Gravitational Potential Energy U = Mgy F

  8. 0 / 100 Which graph of potential energy describes the action of the force in the picture below? • Picture 1 • Picture 2 • Picture 3 To solve, this, note: Cross-Tab Label

  9. Work, Potential Energy, and Mechanical Energy W = - DU = -(Ufinal - Uinitial) and Work-energy theorem W = DK So, W – W = 0 = DK + DU Leads to the Conservation of Mechanical Energy E = K + U E is CONSTANT, SO DE = 0

  10. Conservation of energy can simplify problem solving. Block dropped from height H. What is speed of block just before impact with ground? Ui = MgH Ki = 0 Ki = ½ M v2 Uf = 0 E = constant E = K + U Initial Final H

  11. Path doesn’t matter! Initial Final

  12. Potential energy of a spring If a spring is COMPRESSED or STRETCHED and amount DX from its equilibrium position, it has a stored energy, equal to… This is the same value that you have for the Work done on the spring.

  13. DX M V Pinball shooter An in-class problem solving exercise. A spring is initially compressed by an amount DX by a mass M. The mass is released and slides without friction. Given the spring constant K, the compression distance, and the mass M, what is a formula for the final velocity of the block? How to solve: Write down the initial potential and kinetic energy. Next write down the final potential and kinetic energy. Set them equal (conservation of mechanical energy). Solve for V.

  14. DX M V Pinball shooter: step by step Q: Find a formula for final velocity. 1. Write down initial potential and kinetic energy Initial Potential Energy: U = ½ K DX2 Initial Kinetic Energy: K = ½ M V2 = 0 2. Write down the final potential energy and kinetic energy. 3. Set the initial and final energy to be equal (conservation of energy). 4. Solve for V.

  15. DX M V 0 / 100 Pinball Shooter: The final velocity of the block is given by …… • V = sqrt (2g DX) • V = DX sqrt (K/M) • V = ½ K (DX)^2 GIVEN: DX, M, K, g Cross-Tab Label

  16. DX M V 0 / 100 Pinball Shooter: The final velocity of the block is given by …… Second Chance! Work with your Neighbor. • V = sqrt (2g DX) • V = DX sqrt (K/M) • V = ½ K (DX)^2 GIVEN: DX, M, K, g Cross-Tab Label

  17. From point A to B, spring is pulling, work is negative. F DX From point C to point A, spring is pulling, work is positive. F DX Work done by a spring. C What is the work done BY THE SPRING? Path #1: Tricky part: moving from 4 cm to 2 cm. Path #2: Compare to W=-DU method.

  18. Springs and gravity: potential stored in spring. At equilibrium, Force of spring is equal to force of gravity. KL = Mg Potential energy stored in spring: U = ½ K L2 L M

  19. Potential energy of mass and spring together A different problem from the previous slide. Since only DIFFERENCES in potential energy matter, we can define the ZERO of potential to be the equilibrium position of the mass, after it stretches the spring. Potential of mass: mgy Potential of spring: ½ Ky2 Total: U = mgy + ½ Ky2

  20. Pulleys: solve using energy conservation E init = 0 (my choice of potential) CHECK: Suppose m1 was equal to zero? What if masses are equal?

  21. M H L Combo problem: spring and falling body. Mass M drops from a height L onto a spring loaded platform. How much does the spring compress? Spring constant is K. Use conservation of energy. Write down initial energy of mass and spring. (Be sure to use an easy definition of initial energy.) Write down final energy of mass and spring. This will be at point of maximum compression, when the mass STOPS MOVING!

  22. 0 / 100 M H L How much does the spring compress? • ½ KL^2 = MgL • ½ KL^2 = Mg(L+H) • ½ KL^2 = sqrt(MgH) Solve this equation: Cross-Tab Label

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