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Unformatted text preview: Le (tl8426) HW06 gentle (56245) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The innerspring mattress on your grand mothers bed is held up by 36 vertical springs, each having a spring constant of 6000 N / m. A 31 kg person jumps from a 1 . 96 m platform onto the innersprings. The acceleration of gravity is 9 . 8 m / s 2 . Assume: The springs were initially un stretched and that they stretch equally (typi cal old fashioned bed). Determine the stretch of each of the springs. Correct answer: 0 . 0756721 m. Explanation: We have U s vextendsingle vextendsingle vextendsingle at end + U g vextendsingle vextendsingle vextendsingle end = U g vextendsingle vextendsingle vextendsingle initial . Taking x = 0 at the initial level of the inner springs, this becomes (for n springs) n 1 2 k x 2 mg x = mg h, or n 2 k x 2 mg x mg h = 0 . Let a = n 2 k , b = mg , and c = mg h . Using the quadratic formula x = b b 2 4 a c 2 a (and taking the positive solution to get the lowest position of the person) gives x = mg + radicalbig m 2 g 2 + 2 n k mg h n k . Since m 2 g 2 + 2 n k mg h = (31 kg) 2 (9 . 8 m / s 2 ) 2 + 2 (36) (6000 N / m) (31 kg) (9 . 8 m / s 2 )(1 . 96 m) = 2 . 57326 10 8 N 2 , then x = (31 kg) (9 . 8 m / s 2 ) 36(6000 N / m) + 2 . 57326 10 8 N 2 36(6000 N / m) = 0 . 0756721 m . 002 10.0 points A 3 . 5 kg block is dropped onto a spring of spring constant 1056 N / m from a height of 800 cm. 3 . 5 kg 8 m Find the speed of the block when the com pression of the spring is 19 cm. The accelera tion of gravity is 9 . 81 m / s 2 . Correct answer: 12 . 2391 m / s. Explanation: Let : m = 3 . 5 kg , k = 1056 N / m , h = 800 cm = 8 m , x = 19 cm = 0 . 19 m , and g = 9 . 81 m / s 2 . Applying conservation of energy, U g + U s + K = 0 mg ( h + x ) + 1 2 k x 2 + 1 2 mv 2 = 0 . Since v 2 = 2 g ( h + x ) k x 2 m = 2 (9 . 81 m / s 2 ) (8 m + 0 . 19 m) (1056 N / m) (0 . 19 m) 2 3 . 5 kg = 149 . 796 m 2 / s 2 v = radicalBig 149 . 796 m 2 / s 2 = 12 . 2391 m / s . Le (tl8426) HW06 gentle (56245) 2 keywords: 003 10.0 points The planet Mars has a mass of 6 . 1 10 23 kg and radius of 3 . 5 10 6 m. What is the acceleration of an object in free fall near the surface of Mars? The value of the gravitational constant is 6 . 67259 10 11 N m 2 / kg 2 . Correct answer: 3 . 32268 m / s 2 . Explanation: Let : M = 6 . 1 10 23 kg , R = 3 . 5 10 6 m , and G = 6 . 67259 10 11 N m 2 / kg 2 . Near the surface of Mars, the gravitation force on an object of mass m is F = G M m R 2 , so the acceleration of an object in free fall is a = F m = G M R 2 = (6 . 67259 10 11 N m 2 / kg 2 ) 6 . 1 10 23 kg (3 . 5 10 6 m) 2 = 3 . 32268 m / s 2 ....
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This note was uploaded on 12/05/2011 for the course PHY 301 taught by Professor Swinney during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Swinney
 mechanics

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