This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: alexander (jra2623) homework 15 Turner (92510) 1 This printout should have 13 questions. Multiplechoice 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 d Given : m = 9 kg , = 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 + 0 = 0 + 1 2 mv 2 bottom where L is the length moved along the ramp. v bottom = radicalbig 2 gL sin = 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 = k N = k mg . Applying the workkinetic 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) parenleftbigg 0 + 1 2 mv 2 i parenrightbigg 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 = (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....
View Full
Document
 Summer '08
 Kaplunovsky
 Work

Click to edit the document details