hw11 - khounvivongsy(sk27799 – Homework 11 – Weathers...

Info iconThis preview shows pages 1–2. 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: khounvivongsy (sk27799) – Homework 11 – Weathers – (17104) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A spring with a spring-constant 2 . 1 N / cm is compressed 26 cm and released. The 3 kg mass skids down the frictional incline of height 26 cm and inclined at a 17 ◦ 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 . 2 . 3 kg 17 ◦ μ = . 2 . 7 m 26 cm 26 cm k = 2 . 1 N / cm v f Figure: Not drawn to scale. What is the final velocity v f of the mass? Correct answer: 2 . 68401 m / s. Explanation: Let : g = 9 . 8 m / s 2 = , k = 2 . 1 N / cm = 210 N / m , x = 26 cm = 0 . 26 m , μ = 0 . 2 , ℓ = 0 . 7 m , h = 0 . 26 m , m = 3 kg , and θ = 17 ◦ , 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 (210 N / m) (0 . 26 m) 2 = 7 . 098 J . It gains kinetic energy because of the potential energy lost in moving down the incline K gained = U lost = mg h (2) = (3 kg) (9 . 8 m / s 2 ) (0 . 26 m) = 7 . 644 J . and loses kinetic energy by doing work on the frictional surface K lost = W fr = μmg ℓ cos θ (3) = (0 . 2) (3 kg) (9 . 8 m / s 2 ) × (0 . 7 m) cos(17 ◦ ) = 3 . 93615 J . Since energy is concerved, the final kinetic energy is K f = U s + U l − W fr = (7 . 098 J) + (7 . 644 J) − (3 . 93615 J) = 10 . 8058 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 (10 . 8058 J) (3 kg) = 2 . 68401 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....
View Full Document

This note was uploaded on 02/08/2010 for the course PHYS 1710 taught by Professor Weathers during the Spring '09 term at University of North Carolina Wilmington.

Page1 / 5

hw11 - khounvivongsy(sk27799 – Homework 11 – Weathers...

This preview shows document pages 1 - 2. Sign up to view the full document.

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