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Unformatted text preview: homework 19 – HETTIGER, CHRISTOF – Due: Mar 4 2008, 4:00 am 1 Question 1, chap 8, sect 5. part 1 of 2 10 points A 1410 kg car starts from rest and acceler ates uniformly to 11 . 4 m / s in 10 . 2 s . Assume that air resistance remains con stant at 389 N during this time. Find the average power developed by the engine. Correct answer: 15 . 0132 hp (tolerance ± 1 %). Explanation: m = 1410 kg , v i = 0 m / s , v f = 11 . 4 m / s , and Δ t = 10 . 2 s . The acceleration of the car is a = v f v i Δ t = v f Δ t since v i = 0, so a = 11 . 4 m / s 10 . 2 s = 1 . 11765 m / s 2 . Thus the constant forward force due to the engine is found from summationdisplay F = F engine F air = ma F engine = F air + ma = 389 N + (1410 kg) ( 1 . 11765 m / s 2 ) = 1964 . 88 N . The average velocity of the car during this interval is v av = v f + v i 2 , so the average power output is P = F engine v av = F engine parenleftBig v f 2 parenrightBig = (1964 . 88 N) parenleftbigg 11 . 4 m / s 2 parenrightbiggparenleftbigg 1 hp 764 W parenrightbigg = 15 . 0132 hp . Question 2, chap 8, sect 5. part 2 of 2 10 points Find the instantaneous power output of the engine at t = 10 . 2 s just before the car stops accelerating. Correct answer: 30 . 0264 hp (tolerance ± 1 %). Explanation: The instantaneous velocity is 11 . 4 m / s and the instantaneous power output of the engine is P = F engine v f = (389 N)(11 . 4 m / s) parenleftbigg 1 hp 764 W parenrightbigg = 30 . 0264 hp . Question 3, chap 9, sect 99. part 1 of 2 10 points The universal gravitationan constant is 6 . 6726 × 10 − 11 N m 2 / kg 2 . Three 2 kg masses are located at points in the xy plane as shown in the figure. 39 cm 53 cm What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? Correct answer: 1 . 99553 × 10 − 9 N (tolerance ± 1 %). Explanation: Let : m = 2 kg x = 39 cm , and y = 53 cm . homework 19 – HETTIGER, CHRISTOF – Due: Mar 4 2008, 4:00 am 2 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 mm x 2 = Gm 2 x 2 = (6 . 6726 × 10 − 11 N m 2 / kg 2 )(2 kg) 2 (0 . 39 m) 2 = 1 . 75479 × 10 − 9 N . The other force is pointing in the y direction and has magnitude f 2 = G mm y 2 = Gm 2 y 2 = (6 . 6726 × 10 − 11 N m 2 / kg 2 )(2 kg) 2 (0 . 53 m) 2 = 9 . 50174 × 10 − 10 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 = radicalBig f 2 1 + f 2 2 = bracketleftBig (1 . 75479 × 10 − 9 N) 2 + (9 . 50174 × 10 − 10 N) 2 bracketrightBig 1 2 = 1 . 99553 × 10 − 9 N ....
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This note was uploaded on 05/09/2010 for the course PHYSICS PHY 301k taught by Professor Turner during the Spring '10 term at University of TexasTyler.
 Spring '10
 Turner
 Physics, Resistance, Work

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