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Gravitational potential energy. Conservation of energy h 1 2 Definition: Example: An object of unknown mass is projected with an initial speed, v 0 = 10 m/s at an unknown angle above the horizontal. If air resistance could be neglected, what would be the speed of the object at height, h = 3.3 m above the starting point?

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A. 2 mB. 5 mC. 10 m Example: A box of unknown mass and initial speed v 0 = 10 m/s moves up a frictionless incline angled 30°. How high does the box go before it begins sliding down? Only gravity does work (the normal is perpendicular to the motion), so mechanical energy is conserved. m 30° The really nice thing is, we can apply the same thing to any “incline”

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The really nice thing is, we can apply the same thing to any “incline”: h Turn- around point: where K = 0 E K U v = 0

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h Example: Roller Coaster A roller coaster starts out at the top of a hill of height h. How fast is it going when it reaches the bottom?

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Example: Paul and Kathleen start from rest at the same time on frictionless water slides with different shapes. The only force doing work is gravity mechanical energy is conserved: E i = mgh = E f = 1/2 mv 2 gh = 1/2 v 2 Since they both start from the same height, they have the same velocity at the bottom. b) Who makes it to the bottom first? Even though they both have the same final velocity, Kathleen is at a lower height than Paul for most of her ride. Thus she always has a larger velocity during her ride and therefore arrives earlier! 1) Paul 2) Kathleen 3) both the same a) At the bottom, whose velocity is greater?

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Example: A cart is released from height h in a roller coaster with a loop of radius R. What is the minimum h to keep the cart on the track? h R B A The minimum velocity is fixed by N = 0: The minimum height is given by: mg+N

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θ N mg As v increases, N decreases. When N = 0, the pebble flies off. R Example: A pebble of mass m sits on top of the perfectly spherical head of a snow man. When given a very slight push, it starts sliding down. Where does it leave the snow man’s head? Conservation of energy: 2 equations for v MAX and θ MAX

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Only weight of the pendulum is doing work; weight is a conservative force, so mechanical energy is conserved: L m θ0 θ0 The angle on the other side is also θ 0. θ0 θ0 Example: Pendulum (Conservation of energy)

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The pendulum with a mass of 300 g is deviated from the equilibrium position B to the position A as shown below. Find the speed of the pendulum at the point B after the pendulum is released. A. Energy of the pendulum at the point A: B. Energy of the pendulum at the point B: C. Conservation of energy:

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x x = 0, equilibrium k m μkμk v?v? Example: A 5.00 g block is pushed against a spring with k = 8.00 N/m. The spring is initially compressed 5.00 cm and then released. The coefficient of kinetic friction between the block and the table is What is the speed of the block at x = 0?

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Example: A box sliding on a horizontal frictionless surface runs into a fixed spring and compresses it to a maximum distance x 1 before bouncing back. If the initial speed of box is doubled and its mass if halved, how far (x 2 ) would the spring be compressed? A. x 2 = x 1 B. x 2 = 2 x 1 C. x 2 = 2x 1 x1x1 E f =E i

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Example: Spring 1 is stiffer then spring 2; that is k 1 > k 2. The spring force F of which spring does more work if the springs compressed by the same applied force? F=-kx x=-F/k W=½Fx=½F(-F/k)=-½F 2 /k A. W 1

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Energy Conservation 1. Mechanical energy conservation For closed isolated system 2. Open system 3. Conservative and nonconservative forces Forces such as gravity or the elastic force, for which the work dose not depend on the path taken but only on the initial and final position, are called conservative forces For conservative forces the work done on a closed path (a lop) is equal to zero Friction is a nonconservative force

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Example: A hammer slides along 10 m down a 30 inclined roof and off into the yard, which is 7 m below the roof edge. Right before it hits the ground, its speed is 14.5 m/s. What is the coefficient of kinetic friction between the hammer and the roof? Δx = 10 m h = 7 m v = 14.5 m/s 30 This can be solved using Newton’s laws and kinematics, but it’s looooooooooooooooooooooooong. h’

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