hw9 - Muraj Hamza Homework 9 Due 4:00 am Inst Florin This...

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Muraj, Hamza – Homework 9 – Due: Feb 10 2006, 4:00 am – Inst: Florin 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points In a certain system o± units, 37 . 8 thumbs = 1 meter, 1 pause = 25 . 6 s , and 1 glob = 20 . 5 kg. The unit o± ±orce in this system is a perk (p). How many kiloperks (kp) are in 49 . 5 N? Correct answer: 59 . 8168 kp. Explanation: Basic concept: The use o± conversion ±actors. 1 N = 1 kg · m s 2 1 perk = 1 glob · thumb pause 2 Solution: N = kgm s 2 · 1glob 20 . 5 kg · 37 . 8 thumbs 1m · µ 25 . 6 s 1pause 2 glob · thumb pause 2 · 1kp 1000p kp 002 (part 1 o± 1) 10 points A 4 . 2 g bullet leaves the muzzle o± a ri²e with a speed o± 301 m / s. What total constant ±orce is exerted on the bullet while it is traveling down the 0 . 86 m long barrel o± the ri²e? Correct answer: 221 . 235 N. Explanation: Given : m = 4 . 2 g = 0 . 0042 kg , v = 301 m / s , and Δ x = 0 . 86 m . The acceleration o± the bullet is given by v 2 = v 2 i + 2 a x ) = 2 a x ) since v i = 0. a = v 2 2(Δ x ) and the ±orce exerted on the bullet is X F = ma = m v 2 2(Δ x ) = (0 . 0042 kg)(301 m / s) 2 2(0 . 86 m) = 221 . 235 N . 003 (part 1 o± 1) 10 points At the beginning o± a new school term, a student moves a box o± books by attaching a rope to the box and pulling with a ±orce o± F = 89 . 4 N at an angle o± 63 , as shown in the fgure. The acceleration o± gravity is 9 . 8 m / s 2 . The box o± books has a mass o± 14 kg and the coe³cient o± kinetic ±riction between the bottom o± the box and the ²oor is 0 . 34. 14 kg F 63 μ = 0 . 34 What is the acceleration o± the box? Correct answer: 1 . 50155 m / s 2 . Explanation: Given : m = 14 kg , μ k = 0 . 34 , F = 89 . 4 N , and θ = 63 . Basic Concepts: Vertically there is no acceleration, so F net = X F up - X F down = 0 Horizontally, F net = ma = X F right - X F left
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Muraj, Hamza – Homework 9 – Due: Feb 10 2006, 4:00 am – Inst: Florin 2 Solution Consider the free body diagram for the situation. 14 kg F 63 N mg μmg The tension in the rope can be resolved into a horizontal component of F cos θ and a vertical component of F sin θ . The box is in equilibrium vertically, so F net = N + F sin θ - mg = 0 N = mg - F sin θ . The frictional force is
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hw9 - Muraj Hamza Homework 9 Due 4:00 am Inst Florin This...

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