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Unformatted text preview: rathbun (rar2954) HW #6 Antoniewicz (57380) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 4.0 points Starting from rest, a(n)13 kg block slides 9 m down a frictionless ramp (inclined at 30 from the floor) to the bottom. The block then slides an additional 23 m along the floor before coming to stop. The acceleration of gravity is 9 . 8 m / s 2 . 9 m 13 kg 1 3 k g 30 23 m Find the speed of the block at the bottom of the ramp. Correct answer: 9 . 39149 m / s. Explanation: L m m d Given : m = 13 kg , = 30 , L = 9 m , and d = 23 m . Let P = 0 at the bottom of the ramp. Using conservation of energy along the ramp, we have U i + K i = U f + K f mgL sin + 0 = 0 + 1 2 mv 2 bottom where L is the length moved along the ramp. v bottom = 2 gL sin = 2(9 . 8 m / s 2 )(9 m) sin30 = 9 . 39149 m / s . 002 (part 2 of 3) 3.0 points Find the coefficient of kinetic friction between block and floor. Correct answer: 0 . 195652. Explanation: Let d be the distance moved along the floor. The frictional force is given by f = k N = k mg . Applying the workkinetic energy theorem, we have W f = ( K U g ) f ( K + U g ) i , where K i is the kinetic energy at the bottom of the ramp and K f = 0. k mg d = (0 + 0) 0 + 1 2 mv 2 i k = v 2 i 2 g d = (9 . 39149 m / s) 2 2(9 . 8 m / s 2 )(23 m) = . 195652 . 003 (part 3 of 3) 3.0 points Find the magnitude of the mechanical energy lost due to friction. Correct answer: 573 . 3 J. Explanation: All of the initial potential energy is lost due to friction along the floor, so is mg y i = mg L sin = (13 kg) (9 . 8 m / s 2 ) (9 m) sin30 = 573 . 3 J . rathbun (rar2954) HW #6 Antoniewicz (57380) 2 004 10.0 points The spring has a constant of 29 N / m and the frictional surface is 0 . 4 m long with a coefficient of friction = 1 . 44 . The 4 kg block depresses the spring by 24 cm, then is released. The first drop is 2 . 3 m and the second is 1 m. The acceleration of gravity is 9 . 8 m / s 2 . h h L m k x 1 2 s How far from the bottom of the cliff does it land? Correct answer: 2 . 6422 m. Explanation: Basic concepts The potential energy of a spring is U s = 1 2 k x 2 = 1 2 (29 N / m) (24 cm) 2 = 0 . 8352 J . Gravitational potential energy is U g = mg h = (4 kg) (9 . 8 m / s 2 ) (2 . 3 m) = 90 . 16 J . Work done against friction on a flat surface is W fr = mg L = (1 . 44) (4 kg) (9 . 8 m / s 2 ) (0 . 4 m) = 22 . 5792 J . Kinetic energy is K = 1 2 mv 2 . The height of the cliff determines how long it takes for the mass to reach the bottom. Energy considerations give us the horizon tal velocity as the mass leaves the cliff top....
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This note was uploaded on 03/08/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics

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