HW 05 solution - gutierrez(ig3472 HW5 ditmire(58216 This...

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gutierrez (ig3472) – HW5 – ditmire – (58216) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A spring has a Force constant oF 311 N / m and an unstretched length oF 7 cm. One end is attached to a post that is Free to rotate in the center oF a smooth table, as shown in the top view below. The other end is attached to a 9 kg disk moving in uniForm circular motion on the table, which stretches the spring by 1 cm. Note: ±riction is negligible. 311 N / m 9 kg 8 cm What is the centripetal Force F c on the disk? Correct answer: 3 . 11 N. Explanation: Let : r = 7 cm = 0 . 07 m , Δ r = 1 cm = 0 . 01 m , m = 9 kg , and k = 311 N / m . The centripetal Force is supplied only by the spring. Given the Force constant and the extension oF the spring, we can calculate the Force as F c = k Δ r = (311 N / m) (0 . 01 m) = 3 . 11 N . 002 (part 2 oF 2) 10.0 points What is the work done on the disk by the spring during one Full circle? 1. W = 0 . 00106673 J 2. W = 0 J correct 3. W = 1 . 36785 J 4. W = 0 . 2177 J 5. W = 0 . 683925 J Explanation: Since the Force is always perpendicular to the movement oF the disk, the work done by the spring is zero . 003 (part 1 oF 3) 10.0 points A 1270 kg car accelerates uniFormly From rest to 14 . 2 m / s in 4 . 02 s. ±ind the work done on the car in this time interval. Correct answer: 128 . 041 kJ. Explanation: Given : m = 1270 kg , v f = 14 . 2 m / s , and Δ t = 4 . 02 s . W = Δ K = 1 2 mv 2 f - 0 = 1 2 (1270 kg) (14 . 2 m / s) 2 · 1 kJ 1000 J = 128 . 041 kJ . 004 (part 2 oF 3) 10.0 points ±ind the average power delivered by the en- gine in this time interval. Correct answer: 42 . 6958 hp. Explanation: P = W Δ t
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gutierrez (ig3472) – HW5 – ditmire – (58216) 2 = p 128 . 041 kJ 4 . 02 s Pp 1000 J kJ 1 hp 746 W P = 42 . 6958 hp . 005 (part 3 of 3) 10.0 points Find the instantaneous power delivered by the engine at t = 1 . 84 s. Correct answer: 39 . 0847 hp. Explanation: Given : Δ t 1 = 1 . 84 s In the ±rst 4 . 02 s a = Δ v Δ t = v f - v i Δ t = v f Δ t since v i = 0, so the net force is F = ma = mv f Δ t . At t = 1 . 84 s , the instantaneous velocity is v = v i + a Δ t 1 = a Δ t 1 = v f Δ t Δ t 1 since v i = 0 , and the power output is P = F v = f Δ t 1 · ± v f Δ t Δ t 1 ² = m ± v f Δ t ² 2 Δ t 1 = (1270 kg) p 14 . 2 m / s 4 . 02 s P 2 · (1 . 84 s) p 1 hp 746 W P = 39 . 0847 hp . 006 (part 1 of 5) 10.0 points A crate is pulled up a rough incline. The pulling force is parallel to the incline. The crate is pulled a distance of 5 . 04 m. The acceleration of gravity is 9 . 8 m / s 2 . 9 . 69 kg μ = 0 281 169 N 1 69 m / s 28 . 6 What is the magnitude of the work is done by the gravitational force? Correct answer: 229 . 106 J. Explanation: Let : d = 5 . 04 m , θ = 28 . 6 , m = 9 . 69 kg , g = 9 . 8 m / s 2 , μ = 0 . 281 , and v = 1 . 69 m / s . F μ N N mg v θ The gravitational force directed down the plane is F grav = - sin θ , and the motion is directed up the plane, so the work by the gravity is W grav = - mg d sin θ = - (9 . 69 kg) g (5 . 04 m) sin28 . 6 = - 229 . 106 J | W grav | = 229 . 106 J .
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HW 05 solution - gutierrez(ig3472 HW5 ditmire(58216 This...

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