hw15 - Overton Mays Homework 15 Due 4:00 am Inst Turner...

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

Overton, Mays – Homework 15 – Due: Feb 28 2005, 4:00 am – Inst: Turner 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points A block starts at rest and slides down a ±ric- tionless track except ±or a small rough area on a horizontal section o± the track (as shown in the fgure below). Itleavesthetrackhorizontally, ²iesthrough the air, and subsequently strikes the ground. The acceleration o± gravity is 9 . 81 m / s 2 . μ =0 . 3 1 . 4 m 422 g h 1 . 9m 4 . 16 m 9 81m / s 2 v At what height h above the ground is the block released? Correct answer: 4 . 59705 m. Explanation: Let : g = 9 . 81 m / s 2 , m = 422 g , μ = 0 . 3 , = 1 . 4 m , h = h 1 + h 2 , h 2 = 1 . 9 m , and v x = v . μ =0 . 3 m 4 6m x g 6 . 68 m / s Basic Concepts: Conservation o± Me- chanical Energy U i = U f + K f + W . (1) since v i = 0 m/s. K = 1 2 mv 2 (2) U g = mg h (3) W = μmg ‘. (4) Choosing the point where the block leaves the track as the origin o± the coordinate system, Δ x = v x Δ t (5) h 2 = 1 2 g Δ t 2 (6) since a x i = 0 m/s 2 and v y i = 0 m/s. Solution: From energy conservation Eqs. 1, 2, 3, and 4, we have 1 2 mv 2 x = mg ( h - h 2 ) - μmg ‘ h 1 = v 2 x 2 g + μ‘ (7) h 2 = 1 2 g t 2 (6) x = v x t. (5) Using Eq. 6 and substituting t = x v x ±rom Eq. 5, we have h 2 = 1 2 g µ x v x 2 , so v 2 x = g x 2 2 h 2 . (8) Using Eq. 6 and substituting v 2 x ±rom Eq. 8, we have h = h 2 + h 1 = h 2 + g x 2 2 h 2 2 g + μ‘ = h 2 + x 2 4 h 2 + μ‘ (9) = 1 . 9 m + (4 . 16 m) 2 4(1 . 9 m) + (0 . 3)(1 . 4 m) = 4 . 59705 m .

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

View Full Document
Overton, Mays – Homework 15 – Due: Feb 28 2005, 4:00 am – Inst: Turner 2 002 (part 1 of 3) 10 points A block is pushed against the spring with spring constant k (located on the left-hand side of the track) and compresses the spring a distance 5 . 3 cm from its equilibrium position (as shown in the ±gure below). The block starts at rest, is accelerated by the compressed spring, and slides across a frictionless track except for a small rough area on a horizontal section of the track (as shown in the ±gure below). Itleavesthetrackhorizontally, ²iesthrough the air, and subsequently strikes the ground. The acceleration of gravity is 9 . 81 m / s 2 . μ =0 . 2 1 . 3 m 512 g 2 . 3m 4 . 29 m 9 81m / s v k 5 . 3 cm What is the spring constant k ? Correct answer: 8083
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 7

hw15 - Overton Mays Homework 15 Due 4:00 am Inst Turner...

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

View Full Document
Ask a homework question - tutors are online