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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online