This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 mother’s 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, Δ U g + Δ U s + Δ 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 ....
View
Full Document
 Fall '07
 Swinney
 mechanics, Mass, General Relativity, Correct Answer, Newton's law of universal gravitation

Click to edit the document details