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Unformatted text preview: roofner (bar784) Homework 10 Weathers (17104) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A spring with a springconstant 3 . 4 N / cm is compressed 29 cm and released. The 2 kg mass skids down the frictional incline of height 50 cm and inclined at a 21 angle. The acceleration of gravity is 9 . 8 m / s 2 . The path is frictionless except for a dis tance of 0 . 7 m along the incline which has a coefficient of friction of 0 . 5 . 2 kg 21 = . 5 . 7 m 50 cm 29 cm k = 3 . 4 N / cm v f Figure: Not drawn to scale. What is the final velocity v f of the mass? Correct answer: 4 . 20626 m / s. Explanation: Let : g = 9 . 8 m / s 2 = , k = 3 . 4 N / cm = 340 N / m , x = 29 cm = 0 . 29 m , = 0 . 5 , = 0 . 7 m , h = 0 . 5 m , m = 2 kg , and = 21 , Consider the kinetic energy of the mass. The mass receives its initial kinetic energy from the potential energy of the spring K i = U spring = 1 2 k x 2 (1) = 1 2 (340 N / m) (0 . 29 m) 2 = 14 . 297 J . It gains kinetic energy because of the potential energy lost in moving down the incline K gained = U lost = mg h (2) = (2 kg) (9 . 8 m / s 2 ) (0 . 5 m) = 9 . 8 J . and loses kinetic energy by doing work on the frictional surface K lost = W fr = mg cos (3) = (0 . 5) (2 kg) (9 . 8 m / s 2 ) (0 . 7 m) cos(21 ) = 6 . 40436 J . Since energy is concerved, the final kinetic energy is K f = U s + U l W fr = (14 . 297 J) + (9 . 8 J) (6 . 40436 J) = 17 . 6926 J . However, the final kinetic energy is K f = 1 2 mv 2 . (3) Multiplying by 2 and dividing by m gives us 2 K f m = v 2 , so v = radicalbigg 2 K f m = radicalBigg 2 (17 . 6926 J) (2 kg) = 4 . 20626 m / s . Alternate Explanation: The potential energy at the top of the hill will be converted into kinetic energy at the bottom of the hill minus energy lost due to the nonconservative friction force....
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 Fall '08
 Weathers
 mechanics, Work

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