P150_Exam2_F06sol - Physics 150 1 2 3 4 5 6 7 8 9 10 11 12...

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11. Answer = D 12. Answer = A 13. Answer = C 14. Answer = D 15. Answer = B 16. Answer = C 17. Answer = C 18. Answer = B 19. Answer = D 20. Answer = E Physics 150 Exam #2 Solution 11/6/06 1. Answer = D 2. Answer = B 3. Answer = A 4. Answer = C 5. Answer = E 6. Answer = E 7. Answer = A 8. Answer = A 9. Answer = D 10. Answer = E
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U g U s := U g 263.424mN = h U g mass g := h 1.12 m = This is the height above the release point. height h x d + () := height 0.84 m = This is the height relative to point O. D) U g mass g height := U g 197.568J = This is the potential energy relative to O. As the mass passes point O on the way down, KE U g := v 2KE mass := v 4.058msec -1 = Velocity at O. Physics 150 Exam #2 Solution 11/6/06 Problem #1: x 3.5 cm := mass 24 kg := d 24.5 cm := A) The mass rests on top of the spring and compresses it 3.5 cm. from this we calculate the spring constant. F mass g := F 235.2N = k F x := k 6.72 10 3 × N m = B) Then the spring is compressed an additional distance d. Hence the total compression distance is x + d. The stored energy in the spring is then given by, U s 1 2 k xd + 2 := U s 263.424J = C) When released from rest the mass will rise.
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This note was uploaded on 04/10/2008 for the course PHYSICS 150 taught by Professor Naik during the Winter '08 term at University of Michigan.

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P150_Exam2_F06sol - Physics 150 1 2 3 4 5 6 7 8 9 10 11 12...

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