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Unformatted text preview: homework 04 – ALIBHAI, ZAHID – Due: Feb 14 2007, 4:00 am 1 Question 1 part 1 of 2 10 points Four forces act on a hotair balloon, as shown from the side. 229 N 111 N 216N 80N Note: Figure is not drawn to scale a) Find the magnitude of the resultant force on the balloon. Correct answer: 180 . 056 N (tolerance ± 1 %). Explanation: 229 N 111 N 216 N 80 N 1 8 . 5 6 N Scale: 100 N 49 . 0536 ◦ Basic Concepts: F x,net = F x, 1 + F x, 2 F y,net = F y, 1 + F y, 2 F net = radicalBig ( F x,net ) 2 + ( F y,net ) 2 Given: F x, 1 = 229 N F x, 2 = 111 N F y, 1 = 216 N F y, 2 = 80 N Solution: Consider the horizontal forces: F x,net = 229 N + ( 111 N) = 118 N Consider the vertical forces: F y,net = 216 N + ( 80 N) = 136 N Thus the net force is F net = radicalBig (118 N) 2 + (136 N) 2 = 180 . 056 N . Question 2 part 2 of 2 10 points b) Find the direction of the resultant force (in relation to the 229 N force, with up being positive). Correct answer: 49 . 0536 ◦ (tolerance ± 1 %). Explanation: Basic Concept: tan θ = F y,net F x,net Solution: θ = tan − 1 parenleftbigg F y,net F x,net parenrightbigg = tan − 1 parenleftbigg 136 N 118 N parenrightbigg = 49 . 0536 ◦ . homework 04 – ALIBHAI, ZAHID – Due: Feb 14 2007, 4:00 am 2 Question 3 part 1 of 2 10 points A 9 kg object undergoes an acceleration of 3 . 3 m / s 2 . What is the magnitude of the resultant force acting on it? Correct answer: 29 . 7 N (tolerance ± 1 %). Explanation: Given : m 1 = 9 kg and a 1 = 3 . 3 m / s 2 . F net = m 1 a 1 = (9 kg) ( 3 . 3 m / s 2 ) = 29 . 7 N . Question 4 part 2 of 2 10 points If this same force is applied to a 3 . 9 kg object, what acceleration is produced? Correct answer: 7 . 61538 m / s 2 (tolerance ± 1 %). Explanation: Given : m 2 = 3 . 9 kg . a 2 = F net m 2 = 29 . 7 N 3 . 9 kg = 7 . 61538 m / s 2 . Question 5 part 1 of 2 10 points A car of mass m , initially at rest at time t = 0, is driven to the right as shown along a straight, horizontal road with the engine caus ing a constant force vector F to be applied. While moving, the car encounters a resistance force equal to kvectorv , where vectorv is the velocity of the car and k is a positive constant. Use a dot to represent the center of mass of the car. And on the figure, draw and label vectors to represent all the forces acting on the car as it moves with a velocity vectorv to be the right. car v g O + Determine the horizontal acceleration of the car. 1. a = F + k v m 2. a = F k v m correct 3. a = k v m 4. a = F + k m v 5. a = F k m v Explanation: F k v N mg F net = ma . But F net = F k v , so F k v = ma a = F k v m . Question 6 part 2 of 2 10 points Derive the equation expressing the velocity of the car as a function of time t ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Force, Work

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