HW #6-solutions - rathbun (rar2954) HW #6 Antoniewicz...

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Unformatted text preview: rathbun (rar2954) HW #6 Antoniewicz (57380) 1 This print-out should have 21 questions. Multiple-choice 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 work-kinetic 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.

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HW #6-solutions - rathbun (rar2954) HW #6 Antoniewicz...

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