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Homework 4 Solutions

# Homework 4 Solutions - homework 04 ALIBHAI ZAHID Due 4:00...

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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 hot-air balloon, as shown from the side. 229 N 111 N 216 N 80 N 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 180 . 056 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 .

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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 0 + k v m 2. a = F 0 - k v m correct 3. a = k v m 4. a = F 0 + k m v 5. a = F 0 - k m v Explanation: F 0 k v N mg F net = m a . But F net = F 0 - k v , so F 0 - k v = m a a = F 0 - 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 . 1. v = F 0 k parenleftBig 1 + e k t/m parenrightBig 2. v = F 0 k parenleftBig 1 + e k m/t parenrightBig
homework 04 – ALIBHAI, ZAHID – Due: Feb 14 2007, 4:00 am 3 3. v = F 0 k parenleftBig 1 - e k m/t parenrightBig 4. v = F 0 k parenleftBig e k t/m parenrightBig 5. v = F 0 k parenleftBig 1 - e k t/m parenrightBig correct Explanation: a = d v dt Using the equation from the previous part d v dt = F 0 - k v m (1) Re-arrange and integrating integraldisplay d v F 0 - k v = integraldisplay 1 m dt (2) Let : u = F 0 - k v du = - k dv - 1 k integraldisplay d u u = integraldisplay 1 m dt ln( F 0 - k v ) - ln C = - k m t , where C is a constant.

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