HW13 - homework 13 – PAPAGEORGE, MATT – Due: Feb 19...

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: homework 13 – PAPAGEORGE, MATT – Due: Feb 19 2008, 4:00 am 1 Question 1, chap 6, sect 2. part 1 of 1 10 points A load of 66 N attached to a spring hanging vertically stretches the spring 2 cm. The spring is now placed horizontally on a table and stretched 12 cm. What force is required to stretch it by this amount? Correct answer: 396 N (tolerance ± 1 %). Explanation: Given : F = 66 N , x 1 = 2 cm , and x 2 = 12 cm . The force exerted by the spring is F 1 = k x 1 k = F 1 x 1 . When stretched 12 cm, F 2 = k x 2 = F 1 x 2 x 1 = (66 N) (12 cm) 2 cm = 396 N . Question 2, chap 6, sect 2. part 1 of 2 10 points A 0 . 56 kg mass is attached to a spring with a spring constant 101 N / m so that the mass is allowed to move on a horizontal frictionless surface. The mass is released from rest when the spring is compressed 0 . 137 m. Find the maximum force on the mass. Correct answer: 13 . 837 N (tolerance ± 1 %). Explanation: Basic Concepts: F = ma F = k x The maximum force and acceleration occur at the maximum amplitude of the motion. Solution: The force on the mass is F = k x = (101 N / m) (0 . 137 m) = 13 . 837 N . Question 3, chap 6, sect 2. part 2 of 2 10 points Find the maximum acceleration. Correct answer: 24 . 7089 m / s 2 (tolerance ± 1 %). Explanation: The acceleration is a = F m = 13 . 837 N . 56 kg = 24 . 7089 m / s 2 . Question 4, chap 6, sect 3. part 1 of 2 10 points The coefficient of static friction between the person and the wall is 0 . 75 . The radius of the cylinder is 6 . 12 m . The acceleration of gravity is 9 . 8 m / s 2 . An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away. ω 6 . 12 m What is the minimum angular velocity ω min needed to keep the person from slip- ping downward? Correct answer: 1 . 46119 rad / s (tolerance ± 1 %). Explanation: homework 13 – PAPAGEORGE, MATT – Due: Feb 19 2008, 4:00 am 2 Let : R = 6 . 12 m and μ = 0 . 75 ....
View Full Document

This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas.

Page1 / 5

HW13 - homework 13 – PAPAGEORGE, MATT – Due: Feb 19...

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