025 (part 1 of 2) 10.0 points A potential energy function is given by U = Ax3 +

Question

025 (part 1 of 2) 10.0 points A potential energy function is given by U = Ax3 + B x2 ? C x + D. For positive parameters A, B, C, D this po- tential has two equilibrium positions, one sta- ble and one unstable. Find the stable equilibrium position for A = 2.05 J/m3 , B = 2.45 J/m2 , C = 2.9 J/m, and D = ?3 J . Answer in units of m part 2 026 (part 2 of 2) 10.0 points What is the unstable equilibrium position? Answer in units of m

Explanation / Answer

E = Ax^3 + Bx^2 - Cx + D. E = 2.05x^3 + 2.45x^2 - 2.9x - 3. It is stable when dE/dx is zero. dE/dx = 6.15x^2 + 4.9 x - 2.9 = 0 The maximmum or minimum points are found solving the above equation we get x = 0.3955 m and and x = -1.19225 m ------------------------------------- We find the derivative of dE/dx =6.15x^2 + 4.9 x - 2.9 = 0 Which is 12.3 x +4.9. At x = 0.3955, the value of 12.3 x + 4.9. is positive and At x = -1.19225 m the value of 12.3 x + 4.9. is negative., If f '' (a) > 0 , f has a relative minimum at x = a. Hence the answer is x = 0.3955 m On either side of this positon the potential energy increases and hence at this position the object will be in stable equilibrium. ======================================

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