HW 16 - Morby Grant Homework 16 Due Mar 1 2006 noon Inst...

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Morby, Grant – Homework 16 – Due: Mar 1 2006, noon – Inst: Drummond 1 This print-out should have 11 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 A 1 kg mass slides to the right on a surface having a coefficient of friction 0 . 22 as shown in the figure. The mass has a speed of 3 m / s when contact is made with a spring that has a spring constant 65 N 2 . The mass comes to rest after the spring has been compressed a distance d . The mass is then forced toward the left by the spring and continues to move in that direction beyond the unstretched posi- tion. Finally the mass comes to rest a distance D to the left of the unstretched spring. The acceleration of gravity is 9 . 8 m / s 2 . m k v v v = 0 v = 0 D d i f Find the compressed distance d . Correct answer: 0 . 34041 m. Explanation: The principle we are going to use to solve this problem is that the change in the total energy of the system is equal to the work done by the nonconservative forces W nc = Δ K + Δ U . In our case, the nonconservative force is the frictional force. Therefore W nc = - f d = - (2 . 156 N) d , where the frictional force is f = μ m g = 0 . 22 (1 kg) (9 . 8 m / s 2 ) = 2 . 156 N . The change in kinetic energy is Δ K = 0 - 1 2 m v 2 i = - 1 2 (1 kg) (3 m / s) 2 = - 4 . 5 J , and the change in potential energy is Δ U = 1 2 k d 2 = 1 2 (65 N 2 ) d 2 = (32 . 5 N / m) d 2 , so W nc = Δ K + Δ U - (2 . 156 N) d = - 4 . 5 J + (32 . 5 N / m) d 2 . (32 . 5 N / m) d 2 + (2 . 156 N) d + ( - 4 . 5 J) = 0 . This is a quadratic equation, and since b 2 - 4 ac = (2 . 156 N) 2 - 4(32 . 5 N / m)( - 4 . 5 J) = 589 . 648 N 2 , the positive solution is d = - (2 . 156 N) + 589 . 648 N 2 2 (32 . 5 N / m) = 0 . 34041 m . 002 (part 1 of 3) 10 points A single conservative force F ( x ) = b x + a
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Morby, Grant – Homework 16 – Due: Mar 1 2006, noon – Inst: Drummond 2 acts on a 5 . 14 kg particle, where x is in meters, b = 6 . 52 N / m and a = 3 . 13 N. As the particle moves along the x axis from x 1 = 1 . 45 m to x 2 = 3 . 72 m, calculate the work done by this force.
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