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Unformatted text preview: Physics 2101, Final Exam, Spring 2006 May 11, 2006 Name : Section: (Circle one) 1 (Rupnik, MWF 7:40am) 2 (Rupnik, MWF 9:40am) 3 (Rupnik, MWF 11:40am) 4 (Kirk, MWF 2:40pm) 5 (Kirk, TTh 10:40am) 6 (Gonz´ alez, TTh 1:40pm) • Please be sure to write your name and circle your section above. • For the problems, you must show all your work. Explain your thinking clearly. Lonely right answers will not receive full credit, lonely wrong answers will receive no credit. • For the questions, no work needs to be shown (there is no partial credit). • Please carry units through your calculations when needed, lack of units will result in a loss of points. • You may use scientific or graphing calculators, but you must derive your answer and explain your work. • Feel free to detach, use and keep the formula sheet. No other reference material is allowed during the exam. • GOOD LUCK! 1 Problem 1 (11 points) A 180 g block is dropped from rest onto a relaxed vertical spring that has a spring constant of k = 300 N/m. The block becomes attached to the spring and compresses the spring 12 cm before momentarily stopping. p p p p p p p p p p p p p p p p p p p p p p p (a) (3 pts) Calculate the work done on the block by the gravi tational force from the time the block falls on the spring until the time the block momentarily stops. The work done by the gravitational force F g = mg force is W g = R Δ x F · d s = mg Δ x . Also, the work done by the gravitational force is minus the change in gravitational potential energy. The change in gravitational energy is negative, since the mass moves down a distance Δ x : Δ U g = mg Δ x = . 18kg × 9 . 8m / s 2 × . 12m = . 21J, and thus W g = +0 . 21 J . (b) (4 pts) Calculate work done on the block by the spring force from the time the block falls on the spring until the time the block momentarily stops. The work done by the spring force F s = kx force is W s = R  Deltax F · d s = (1 / 2) k (Δ x ) 2 . The work done by the spring is minus the change in spring poten tial energy. Initially, the spring is not compressed or stretchesd and the potential energy is zero; when the block momentarily stops, the spring potential energy is (1 / 2) k (Δ x ) 2 . The change in potential energy is pos itive: Δ U s = (1 / 2) k (Δ x ) 2 , and the work done by the spring (which is negative) is W s = (1 / 2) k (Δ x ) 2 = . 5 × 300N / m × (0 . 12m) 2 = 2 . 16J. (c) (4 pts) Calculate the height relative to the fully compressed spring from which the block was initially dropped. The kinetic energy of the block is zero at the time where it momentarily stops, and it is also zero at the time when it is released from rest at a distance H from the lowest point. Thus, the change in potential energy is zero: 0 = Δ K + Δ U = 0 + Δ U = Δ U g + Δ U s = mgy + (1 / 2) k (Δ x ) 2 = 0 y = k (Δ x ) 2 2 mg = 300N / m × (0 . 12m) 2 2 × . 18kg × 9 . 8m / s 2 = 1 . 22m 2 Problem 2 (11 points) The figure shows a block with mass m and initial speed...
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This note was uploaded on 05/20/2011 for the course PHYS 2101 taught by Professor Grouptest during the Spring '07 term at LSU.
 Spring '07
 GROUPTEST
 Physics

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