Homework 16

# Homework 16 - homework 16 – BAUTISTA ALDO – Due Mar 1...

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Unformatted text preview: homework 16 – BAUTISTA, ALDO – Due: Mar 1 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 3. 10 points. Starting from rest, a 11.0 kg suitcase slides 3.00 m down a frictionless ramp inclined at 31 . ◦ from the floor. The suitcase then slides an additional 5.00 m along the floor before coming to a stop. The acceleration of gravity is 9 . 81 m / s 2 . a) Find the speed of the suitcase at the bottom of the ramp. Correct answer: 5 . 50592 m / s (tolerance ± 1 %). Explanation: Basic Concepts: For the slide down the ramp, mechanical energy is conserved. U i = K f since v i = 0 m/s and h f = 0 m. U g = mgh K = 1 2 mv 2 h i = d 1 sin θ Given: m = 11 . 0 kg d 1 = 3 . 00 m θ = 31 . ◦ g = 9 . 81 m / s 2 Solution: mgh i = 1 2 mv f 2 v f = p 2 gh i = p 2 g d 1 sin θ = q 2(9 . 81 m / s 2 )(3 m) sin 31 ◦ = 5 . 50592 m / s Question 2 Part 2 of 3. 10 points. b) Find the coefficient of kinetic friction between the suitcase and the floor. Correct answer: 0 . 309023 (tolerance ± 1 %). Explanation: Basic Concepts: For the horizontal slide across the floor, W net = Δ K = K f- K i =- K i since K f = 0 J. W net = F k d 2 cos θ =- F k d 2 since θ = 180 ◦ ⇒ cos θ =- 1. F k = μ k mg The K i of the horizontal slide equals K f = U i of the slide down the ramp. W net =- U i =- mgh i Given: d 2 = 5 . 00 m Solution:- μ k mgd 2 =- mgh i =- mgd 1 sin θ μ k = d 1 sin θ d 2 = (3 m) sin 31 ◦ 5 m = 0 . 309023 Question 3 Part 3 of 3. 10 points. c) Find the change in mechanical energy due to friction. Correct answer:- 166 . 733 J (tolerance ± 1 %). Explanation: Basic Concept: Δ ME = K f- U i =- U i since K f = 0 J. homework 16 – BAUTISTA, ALDO – Due: Mar 1 2006, 4:00 am 2 Solution: Δ ME =- mgh i =- mgd 1 sin θ =- (11 kg)(9 . 81 m / s 2 )(3 m) · sin 31 ◦ =- 166 . 733 J Question 4 Part 1 of 2. 10 points. A(n) 1700 g block is pushed by an external force against a spring (with a 22 N / cm spring constant) until the spring is compressed by 14 cm from its uncompressed length. The compressed spring and block rests at the bot- tom of an incline of 37 ◦ . The acceleration of gravity is 9 . 8 m / s 2 . Note: The spring lies along the surface of the ramp (see figure). Assume: The ramp is frictionless....
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Homework 16 - homework 16 – BAUTISTA ALDO – Due Mar 1...

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