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Unformatted text preview: HW 8 Solutions 1a. The spring constant k is given by F/x = 800 / . 2 = 4000N / m. The potential energy is therefore U = 1 2 kx 2 = 80J. 1b. U = 1 2 kx 2 = 1 2 (4000 N m )(0 . 05m) 2 = 5J. 2a. The set of points { r  ≤ r ≤ 2fm } are neutral equilibria. The point r = 4fm is an unstable equilibrium. 2b. The magnitude of the force is maximized when  dU/dx  is maximized at the point r = 3fm. 2c. The nucleus is at r = and F = dU/dx so the force is attractive when dU/dx > 0, for { r  2fm < r < 4fm } , and repulsive when dU/dx < 0, for { r  2fm < r < 4fm } 2e. U → 0 for large r , so the particle can go as far as r = 5fm, where U = 1fJ. 2f. The maximum U is 2 fJ, which is 4 fJ greater than U inside the nucleus, so the particle need an additional 3 fJ of kinetic energy. 2g. U max U ∞ = 2fJ of kinetic energy. 3a. By conservation of energy, mg (4) sin θ = 1 2 mv 2 + μ k mg (4) cos θ , so v = (2(9 . 8)(4) sin 53 . 1 2(9 . 8)(4)(0 . 2) cos 53 . 1) 1 / 2 = 7 . 30m / s....
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 Spring '07
 LECLAIR,A
 mechanics, Energy, Force, Kinetic Energy, Potential Energy, HW 8 Solutions, fJ

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