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min (mm64262) – Homework 3 – lee – (GEDB00864) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points 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 10.0points 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 . 13 kg μ = 0 . 287 150 N 1 . 29 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

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min (mm64262) – Homework 3 – lee – (GEDB00864) 2 v = 1 . 29 m / s . F μ N N m g v θ The work-energy 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 - m g cos θ = 0 ⇒ N = m g cos θ . Thus W fric = - μ m g d cos θ = - (0 . 287) (13 kg) (9 . 8 m / s 2 ) × (5 . 21 m) cos 21 = - 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 = - m g d sin θ = - (13 kg) (9 . 8 m / s 2 ) × (5 . 21 m) sin 21 = - 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(part1of2)10.0points 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. 819 N / m 1 kg 8 cm What is the centripetal force F c on the disk? Correct answer: 16 . 38 N. Explanation: Let : r = 6 cm = 0 . 06 m , Δ r = 2 cm = 0 . 02 m , m = 1 kg , and k = 819 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 = (819 N / m) (0 . 02 m) = 16 . 38 N .
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