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Unformatted text preview: min (mm64262) – Homework 3 – lee – (GEDB00864) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The sketch shows a coin at the edge of a turntable. The weight of the coin is shown by the vector W . Two other forces act on the coin: the normal force and a force of friction that prevents it from sliding off the edge. ω W Draw the two remaining force fectors. 1. f N 2. f N correct 3. f N 4. None of these graphs is correct. 5. f N 6. f N 7. f N 8. f N 9. f N Explanation: f W N The friction force must point inward in or der to keep the coin moving in a circle (other wise it will fly off the edge of the disk), and the normal force counteracts the force of gravity. 002 10.0 points A crate is pulled by a force (parallel to the incline) up a rough incline. The crate has an initial speed shown in the figure below. The crate is pulled a distance of 5 . 21 m on the incline by a 150 N force. The acceleration of gravity is 9 . 8 m / s 2 . 1 3 k g μ = . 2 8 7 1 5 N 1 . 2 9 m / s 21 ◦ What is the change in kinetic energy of the crate? Correct answer: 365 . 787 J. Explanation: Let : F = 150 N , d = 5 . 21 m , θ = 21 ◦ , m = 13 kg , g = 9 . 8 m / s 2 , μ = 0 . 287 , and min (mm64262) – Homework 3 – lee – (GEDB00864) 2 v = 1 . 29 m / s . F μN N mg v θ The workenergy theorem with nonconser vative forces reads W fric + W appl + W gravity = Δ K To find the work done by friction we need the normal force on the block from Newton’s law summationdisplay F y = N  mg cos θ = 0 ⇒ N = mg cos θ . Thus W fric = μmg d cos θ = (0 . 287) (13 kg) (9 . 8 m / s 2 ) × (5 . 21 m) cos21 ◦ = 177 . 845 J . The work due to the applied force is W appl = F d = (150 N) (5 . 21 m) = 781 . 5 J , and the work due to gravity is W grav = mg d sin θ = (13 kg) (9 . 8 m / s 2 ) × (5 . 21 m) sin21 ◦ = 237 . 868 J , so that Δ K = W fric + W appl + W grav = ( 177 . 845 J) + (781 . 5 J) + ( 237 . 868 J) = 365 . 787 J . 003 (part 1 of 2) 10.0 points A spring has a force constant of 819 N / m and an unstretched length of 6 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 1 kg disk moving in uniform circular motion on the table, which stretches the spring by 2 cm. Note: Friction is negligible....
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This note was uploaded on 05/24/2011 for the course ENGINEERIN 1 taught by Professor Lee during the Spring '11 term at Sungkyunkwan.
 Spring '11
 LEE

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