This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **homework 16 – PAPAGEORGE, MATT – Due: Feb 26 2008, 4:00 am 1 Question 1, chap 8, sect 1. part 1 of 1 10 points A 7 . 3 kg mass is attached to a light cord that passes over a massless, frictionless pulley. The other end of the cord is attached to a 2 . 3 kg mass. The acceleration of gravity is 9 . 8 m / s 2 . 7 . 3 m ω 7 . 3 kg 2 . 3 kg Use conservation of energy to determine the final speed of the first mass after it has fallen (starting from rest) 7 . 3 m. Correct answer: 8 . 63255 m / s (tolerance ± 1 %). Explanation: Let : m 1 = 7 . 3 kg , m 2 = 2 . 3 kg , and ℓ = 7 . 3 m . Consider the free body diagrams 7 . 3 kg 2 . 3 kg T T m 1 g m 2 g a a Let the figure represent the initial config- uration of the pulley system (before m 1 falls down). From the conservation of energy K i + U i = K f + U f 0 + m 1 g ℓ = m 2 g ℓ + 1 2 m 1 v 2 + 1 2 m 2 v 2 ( m 1- m 2 ) g ℓ = 1 2 ( m 1 + m 2 ) v 2 Therefore v = radicalBigg ( m 1- m 2 ) ( m 1 + m 2 ) 2 g ℓ = bracketleftbigg 7 . 3 kg- 2 . 3 kg 7 . 3 kg + 2 . 3 kg × 2 (9 . 8 m / s 2 )(7 . 3 m) bracketrightbigg 1 / 2 = 8 . 63255 m / s . Question 2, chap 8, sect 1. part 1 of 3 10 points A block starts at rest and slides down a fric- tionless track (as shown in the figure below). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9 . 81 m / s 2 . b b b b b b b b b b b b 403 g 5 . 3m 2 . 2m x 9 . 81m / s 2 v What is the speed v of the ball when it leaves the track? Correct answer: 7 . 79885 m / s (tolerance ± 1 %). Explanation: Let : x = 5 . 22303 m , g = 9 . 81 m / s 2 , m = 403 g , h 1 =- 3 . 1 m , h 2 =- 2 . 2 m , h = h 1 + h 2 , and v x = v . homework 16 – PAPAGEORGE, MATT – Due: Feb 26 2008, 4:00 am 2 b b b b b b b b b b b b m h h 1 h 2 5 . 2 m g v Basic Concepts: Conservation of 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 ℓ. (4) Choosing the point where the block leaves the track as the origin of 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....

View
Full
Document