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physics HW6

# physics HW6 - rose(kmr2427 HW#6 Antoniewicz(57380 This...

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rose (kmr2427) – 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(part1of3)4.0points Starting from rest, a(n)8 kg block slides 13 . 5 m down a frictionless ramp (inclined at 30 from the floor) to the bottom. The block then slides an additional 28 . 5 m along the floor before coming to stop. The acceleration of gravity is 9 . 8 m / s 2 . 13 . 5 m 8 kg 8 kg 30 28 . 5 m Find the speed of the block at the bottom of the ramp. Correct answer: 11 . 5022 m / s. Explanation: L m m θ d Given : m = 8 kg , θ = 30 , L = 13 . 5 m , and d = 28 . 5 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 θ + 0 = 0 + 1 2 mv 2 bottom where L is the length moved along the ramp. v bottom = radicalbig 2 gL sin θ = radicalBig 2(9 . 8 m / s 2 )(13 . 5 m) sin 30 = 11 . 5022 m / s . 002(part2of3)3.0points Find the coefficient of kinetic friction between block and floor. Correct answer: 0 . 236842. Explanation: Let d be the distance moved along the floor. The frictional force is given by f = μ k N = μ 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. μ k m g d = (0 + 0) parenleftbigg 0 + 1 2 m v 2 i parenrightbigg μ k = v 2 i 2 g d = (11 . 5022 m / s) 2 2(9 . 8 m / s 2 )(28 . 5 m) = 0 . 236842 . 003(part3of3)3.0points Find the magnitude of the mechanical energy lost due to friction. Correct answer: 529 . 2 J. Explanation: All of the initial potential energy is lost due to friction along the floor, so is m g y i = m g L sin θ = (8 kg) (9 . 8 m / s 2 ) (13 . 5 m) sin 30 = 529 . 2 J .

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rose (kmr2427) – HW #6 – Antoniewicz – (57380) 2 004 10.0points The spring has a constant of 10 N / m and the frictional surface is 0 . 4 m long with a coefficient of friction μ = 2 . 24 . The 5 kg block depresses the spring by 28 cm, then is released. The first drop is 3 m and the second is 2 m. The acceleration of gravity is 9 . 8 m / s 2 . μ h h L m k x 1 2 s How far from the bottom of the cliff does it land? Correct answer: 4 . 11047 m. Explanation: Basicconcepts The potential energy of a spring is U s = 1 2 k x 2 = 1 2 (10 N / m) (28 cm) 2 = 0 . 392 J . Gravitational potential energy is U g = m g h = (5 kg) (9 . 8 m / s 2 ) (3 m) = 147 J . Work done against friction on a flat surface is W fr = μ m g L = (2 . 24) (5 kg) (9 . 8 m / s 2 ) (0 . 4 m) = 43 . 904 J . Kinetic energy is K = 1 2 m v 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. The mass gets its initial kinetic energy from the spring, then it gains kinetic energy by dropping a distance h 1 , then it loses kinetic energy by doing work against friction. U s + U g W fr = K f = 1 2 m v 2 Since v 2 = 2 ( U s + U g W fr ) m = 2 (0 . 392 J + 147 J 43 . 904 J) 5 kg = 41 . 3952 m 2 / s 2 then v = radicalBig 41 . 3952 m 2 / s 2 = 6 . 43391 m / s .
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