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HW #6-solutions

# HW #6-solutions - rathbun(rar2954 HW#6 Antoniewicz(57380...

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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 13 kg 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 θ d Given : m = 13 kg , θ = 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 θ + 0 = 0 + 1 2 mv 2 bottom where L is the length moved along the ramp. v bottom = 2 gL sin θ = 2(9 . 8 m / s 2 )(9 m) sin 30 = 9 . 39149 m / s . 002 (part 2 of 3) 3.0 points Find the coe ffi cient 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 = 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) - 0 + 1 2 m v 2 i μ k = v 2 i 2 g d = (9 . 39149 m / s) 2 2(9 . 8 m / s 2 )(23 m) = 0 . 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 m g y i = m g L sin θ = (13 kg) (9 . 8 m / s 2 ) (9 m) sin 30 = 573 . 3 J .

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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 coe ffi cient of friction μ = 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 . μ h h L m k x 1 2 s How far from the bottom of the cli ff 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 = m g 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 = μ m g L = (1 . 44) (4 kg) (9 . 8 m / s 2 ) (0 . 4 m) = 22 . 5792 J . Kinetic energy is K = 1 2 m v 2 . The height of the cli ff 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 cli ff 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 . 8352 J + 90 . 16 J - 22 . 5792 J) 4 kg = 34 . 208 m 2 / s 2 then v = 34 . 208 m 2 / s 2 = 5 . 84876 m / s .
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HW #6-solutions - rathbun(rar2954 HW#6 Antoniewicz(57380...

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