HW06-solutions

# HW06-solutions - lee(cyl327 – HW06 – gentle –(56245 1...

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Unformatted text preview: lee (cyl327) – HW06 – gentle – (56245) 1 This print-out should have 16 questions. Multiple-choice 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 20 vertical springs, each having a spring constant of 5000 N / m. A 32 kg person jumps from a 1 . 99 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 . 1149 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 = (32 kg) 2 (9 . 8 m / s 2 ) 2 + 2 (20) (5000 N / m) (32 kg) × (9 . 8 m / s 2 )(1 . 99 m) = 1 . 24911 × 10 8 N 2 , then x = (32 kg) (9 . 8 m / s 2 ) 20(5000 N / m) + √ 1 . 24911 × 10 8 N 2 20(5000 N / m) = 0 . 1149 m . 002 10.0 points A 9 . 2 kg block is dropped onto a spring of spring constant 3723 N / m from a height of 700 cm. 9 . 2 kg 7 m Find the speed of the block when the com- pression of the spring is 12 cm. The accelera- tion of gravity is 9 . 81 m / s 2 . Correct answer: 11 . 5701 m / s. Explanation: Let : m = 9 . 2 kg , k = 3723 N / m , h = 700 cm = 7 m , x = 12 cm = 0 . 12 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 ) (7 m + 0 . 12 m)- (3723 N / m) (0 . 12 m) 2 9 . 2 kg = 133 . 867 m 2 / s 2 v = radicalBig 133 . 867 m 2 / s 2 = 11 . 5701 m / s . lee (cyl327) – 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 . 4 × 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 . 521 m / s 2 . Explanation: Let : M = 6 . 1 × 10 23 kg , R = 3 . 4 × 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 . 4 × 10 6 m) 2 = 3 . 521 m / s 2 ....
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## This note was uploaded on 02/28/2012 for the course PHY 316 taught by Professor Markert during the Fall '10 term at University of Texas.

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HW06-solutions - lee(cyl327 – HW06 – gentle –(56245 1...

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