Physics hw 6 solutions

Physics hw 6 solutions - mata(jpm2873 – HW#6 –...

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Unformatted text preview: mata (jpm2873) – HW #6 – Antoniewicz – (57420) 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)8 kg block slides 11 . 5 m down a frictionless ramp (inclined at 30 ◦ from the floor) to the bottom. The block then slides an additional 25 . 5 m along the floor before coming to stop. The acceleration of gravity is 9 . 8 m / s 2 . 1 1 . 5 m 8 kg 8 k g 30 ◦ 25 . 5 m Find the speed of the block at the bottom of the ramp. Correct answer: 10 . 616 m / s. Explanation: L m m θ d Given : m = 8 kg , θ = 30 ◦ , L = 11 . 5 m , and d = 25 . 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 )(11 . 5 m) sin 30 ◦ = 10 . 616 m / s . 002 (part 2 of 3) 3.0 points Find the coefficient of kinetic friction between block and floor. Correct answer: 0 . 22549. 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 mg d = (0 + 0) − parenleftbigg 0 + 1 2 mv 2 i parenrightbigg μ k = v 2 i 2 g d = (10 . 616 m / s) 2 2(9 . 8 m / s 2 )(25 . 5 m) = . 22549 . 003 (part 3 of 3) 3.0 points Find the magnitude of the mechanical energy lost due to friction. Correct answer: 450 . 8 J. Explanation: All of the initial potential energy is lost due to friction along the floor, so is mg y i = mg L sin θ = (8 kg) (9 . 8 m / s 2 ) (11 . 5 m) sin30 ◦ = 450 . 8 J . mata (jpm2873) – HW #6 – Antoniewicz – (57420) 2 004 10.0 points The spring has a constant of 27 N / m and the frictional surface is 0 . 2 m long with a coefficient of friction μ = 1 . 08 . The 6 kg block depresses the spring by 15 cm, then is released. The first drop is 1 . 7 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: 3 . 45157 m. Explanation: Basic concepts The potential energy of a spring is U s = 1 2 k x 2 = 1 2 (27 N / m) (15 cm) 2 = 0 . 30375 J . Gravitational potential energy is U g = mg h = (6 kg) (9 . 8 m / s 2 ) (1 . 7 m) = 99 . 96 J . Work done against friction on a flat surface is W fr = μmg L = (1 . 08) (6 kg) (9 . 8 m / s 2 ) (0 . 2 m) = 12 . 7008 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....
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Physics hw 6 solutions - mata(jpm2873 – HW#6 –...

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