Solution to Part A: Algebraic Solution: is conserved for all the same reasons as above: ME Uf + Kf = Ui + Ki Uf + 0 = 0 + (2 (¼by4 Mgy) + 0 = 0 + (2 → → – J) J) ¼(4 (20 = 2 (1 (20 2 = 0 – → – – N/m3)y4 N)y J N/m3)y4 N)y Nm So that’s the bad way to do the problem. I don’t know how to that 4th order polynomial. This solver know definitions strongly enough to recognize the gifts that were given via the graph. Graphical Solution: is conserved for all the same reasons as ME Uf + Kf = Ui + Ki Uf + 0 = 0 + (2 → J)
Solution to Part A: Algebraic Solution: ME is conserved for all the same reasons as above: Uf + Kf = Ui + Ki → Uf + 0 = 0 + (2 J) → (¼by4 – Mgy) + 0 = 0 + (2 J) ¼(4 N/m3)y4 – (20 N)y = 2 J → (1 N/m3)y4 – (20 N)y – 2 Nm = 0 So that’s the bad way to do the problem. I don’t know how to that 4th order polynomial. This solver know definitions strongly enough to recognize the gifts that were given via the graph. Graphical Solution: ME is conserved for all the same reasons as Uf + Kf = Ui + Ki → Uf + 0 = 0 + (2 J)