# hw15 - alexander(jra2623 – homework 15 – Turner...

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Unformatted text preview: alexander (jra2623) – homework 15 – Turner – (92510) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Starting from rest, a(n)9 kg block slides 12 . 9 m down a frictionless ramp (inclined at 30 ◦ from the floor) to the bottom. The block then slides an additional 10 . 1 m along the floor before coming to stop. The acceleration of gravity is 9 . 8 m / s 2 . 1 2 . 9 m 9 kg 9 k g 30 ◦ 10 . 1 m Find the speed of the block at the bottom of the ramp. Correct answer: 11 . 2437 m / s. Explanation: L m m θ d Given : m = 9 kg , θ = 30 ◦ , L = 12 . 9 m , and d = 10 . 1 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 )(12 . 9 m) sin30 ◦ = 11 . 2437 m / s . 002 (part 2 of 3) 10.0 points Find the coefficient of kinetic friction between block and floor. Correct answer: 0 . 638614. 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 = (11 . 2437 m / s) 2 2(9 . 8 m / s 2 )(10 . 1 m) = . 638614 . 003 (part 3 of 3) 10.0 points Find the magnitude of the mechanical energy lost due to friction. Correct answer: 568 . 89 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 = (9 kg) (9 . 8 m / s 2 ) (12 . 9 m) sin30 ◦ = 568 . 89 J . alexander (jra2623) – homework 15 – Turner – (92510) 2 004 (part 1 of 2) 10.0 points An 94 kg skydiver jumps out of an airplane at an altitude of 1123 m and opens the parachute at an altitude of 142 m. The total retardingat an altitude of 142 m....
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hw15 - alexander(jra2623 – homework 15 – Turner...

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