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Hw 3 Solutions

# Hw 3 Solutions - Bell Evan Homework 3 Due Feb 2 2005 11:00...

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Bell, Evan – Homework 3 – Due: Feb 2 2005, 11:00 pm – Inst: Furlan 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Sue and Jenny kick a soccer ball at exactly the same time. Sue’s foot exerts a force of 65 . 9 N to the north. Jenny’s foot exerts a force of 98 . 1 N to the east. a) What is the magnitude of the resultant force on the ball? Correct answer: 118 . 18 N. Explanation: f j F f s θ The magnitude of the resultant can be cal- culated using the Pythagorean Theorem: F = q f 2 s + f 2 j = q (65 . 9 N) 2 + (98 . 1 N) 2 = 118 . 18 N 002 (part 2 of 2) 10 points b) What is the direction of the resultant force (measured from East)? Correct answer: 33 . 8918 . Explanation: Jenny’s force is directed east, so f s is the side opposite θ and f j is the side adjacent, and tan θ = f s f j θ = arctan f s f j = arctan 65 . 9 N 98 . 1 N = 33 . 8918 003 (part 1 of 1) 10 points Beth, a construction worker, attempts to pull a stake out of the ground by pulling on a rope that is attached to the stake. The rope makes an angle of 59 . 9 with the horizontal. If Beth exerts a force of 137 . 2 N on the rope, what is the magnitude of the upward component of the force acting on the stake? Correct answer: 118 . 699 N. Explanation: θ θ F F f v θ is the angle the rope makes with the horizontal, so the upward component of the force is the side opposite θ , and sin θ = f v F f v = F sin θ = (137 . 2 N) sin 59 . 9 = 118 . 699 N 004 (part 1 of 1) 10 points A force is applied to a 2 . 2 kg mass and pro- duces 2 . 2 m / s 2 acceleration. What acceleration would be produced by the same force applied to a 7 . 9 kg mass? Correct answer: 0 . 612658 m / s 2 . Explanation: The force acting on both masses is the same, so F = m 1 a 1 = m 2 a 2 a 2 = m 1 a 1 m 2 = (2 . 2 kg) (2 . 2 m / s 2 ) 7 . 9 kg = 0 . 612658 m / s 2 005 (part 1 of 1) 10 points A person weighing 0 . 8 kN rides in an elevator that has a downward acceleration of 1 . 9 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 .

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Bell, Evan – Homework 3 – Due: Feb 2 2005, 11:00 pm – Inst: Furlan 2 What is the magnitude of the force of the elevator floor on the person? Correct answer: 0 . 644898 kN. Explanation: Basic Concepts: X ~ F = m~a ~ W = m~g Solution: Since W = m g , F net = ma = W - f f = W - ma = W 1 - a g = (0 . 8 kN) 1 - 1 . 9 m / s 2 9 . 8 m / s 2 = 0 . 644898 kN . 006 (part 1 of 3) 10 points The following 3 questions refer to a toy car which is given a quick push so that it rolls up an inclined ramp. After it is released, it rolls up, reaches its highest point and rolls back down again. Friction is so small it can be ignored. Indicate the net force acting on the car for each of the cases described below. The car is moving up the ramp after it is released. 1. Net constant force down the ramp. cor- rect 2. Net decreasing force up the ramp.
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Hw 3 Solutions - Bell Evan Homework 3 Due Feb 2 2005 11:00...

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