Homework 18 - Gozick Brandon Homework 18 Due 5:00 pm Inst D...

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Gozick, Brandon – Homework 18 – Due: Nov 17 2006, 5:00 pm – Inst: D Weathers 1 This print-out should have 16 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 1) 10 points Given: G = 6 . 6726 × 10 - 11 N m 2 / kg 2 Three 2 kg masses are located at points in the x-y plane as shown in the figure. 36 cm 50 cm What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? Correct answer: 2 . 31972 × 10 - 9 N. Explanation: Let : m = 2 kg x = 36 cm , and y = 50 cm . Basic Concepts: Newton’s Law of Grav- itation: F g = G m 1 m 2 r 2 We calculate the forces one by one, and then add them using the superposition prin- ciple. The force from the mass on the right is pointing in the x direction, and has magni- tude f 1 = G m m x 2 = G m 2 x 2 = (6 . 6726 × 10 - 11 N m 2 / kg 2 )(2 kg) 2 (0 . 36 m) 2 = 2 . 05944 × 10 - 9 N . The other force is pointing in the y direction and has magnitude f 2 = G m m y 2 = G m 2 y 2 = (6 . 6726 × 10 - 11 N m 2 / kg 2 )(2 kg) 2 (0 . 5 m) 2 = 1 . 06762 × 10 - 9 N . f 2 f 1 F θ Now we simply add the two forces, using vector addition. Since they are at right angles to each other, however, we can use Pythago- ras’ theorem as well: F = q f 2 1 + f 2 2 = h (2 . 05944 × 10 - 9 N) 2 + (1 . 06762 × 10 - 9 N) 2 i 1 2 = 2 . 31972 × 10 - 9 N . As you see, this force is very small. If the masses in the picture are standing on a table (the view being from above), typically the force of static friction will not be overcome. In agreement with common sense, then, the masses will not move. Note: The angle θ shown in the figure is θ = arctan f 2 f 1 = arctan (1 . 06762 × 10 - 9 N) (2 . 05944 × 10 - 9 N) = 27 . 4022 .
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Gozick, Brandon – Homework 18 – Due: Nov 17 2006, 5:00 pm – Inst: D Weathers 2 keywords: 002 (part 1 of 1) 10 points A m = 49 . 3 kg object is released from rest at a distance h = 1 . 08377 R above the Earth’s surface. The acceleration of gravity is 9 . 8 m / s 2 . For the Earth, R E = 6 . 38 × 10 6 m, M = 5 . 98 × 10 24 kg. The gravitational acceleration at the surface of the earth is g = 9 . 8 m / s 2 . Find the speed of the object when it strikes the Earth’s surface. Neglect any atmospheric friction. Caution: You must take into account that the gravitational acceleration depends on dis- tance between the object and the center of the earth. Correct answer: 8064 . 59 m / s. Explanation: Let : h R E = 1 . 08377 , or h = (1 . 08377) R E , where R E = 6 . 38 × 10 6 m , M E = 5 . 98 × 10 24 kg , g = 9 . 8 m / s 2 , and m = 49 . 3 kg , not required . Basic Concepts: U = - G M m r and g = G M R 2 E . Apply conservation of energy from the point where the mass is released to the Earth’s surface U 1 + K 1 = U 2 + K 2 - G m M E R E + h + 0 = - G m M E R E + 1 2 m v 2 . Solving for v gives v 2 = 2 G M E 1 R E - 1 R E + h = 2 G M E R E 1 - 1 1 + h R E = 2 G M E R 2 E h 1 + h R E = 2 g h 1 + h R E , or v = v u u u t 2 g h 1 + h R E = s 2 (9 . 8 m / s 2 ) (1 . 08377) (6 . 38 × 10 6 m) 1 + (1 . 08377) = 8064 . 59 m / s .
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