Practice_Final_Solutions

Practice_Final_Solutions - Solutions to Phys. 6A Practice...

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Solutions to Phys. 6A Practice Final Derek Schaeffer 1) Kinematics a. h max = 2296 m b. v max = v x = v 0 cos(θ) = 212 m/s [at top of trajectory, v y = 0] c. a max = -g = -9.8 m/s 2 [a is always –g] d. t = 2* v y / g = 43.2 s e. d = v 0 cos(θ)t = 9184 m 2) Forces a. μ k = (F – m 1 a 1 ) / (m 1 g) = 0.24 b. a 2 = ƒ k / m 2 = 1.4 m/s 2 Note: the friction force between the two blocks is towards the left for the top block and towards the right (equal and opposite) for the bottom block) 3) Radial Forces a. a r = v 2 / L = 2.45 m/s 2 b. a t = gsin(θ) = 2.54 m/s 2 [sum over forces in the tangential direction] c. T = ma r + mgcos(θ) = 11.9 N 4) Energy and Radial Forces a. v top = √(2g (H – 2R)) = 14.7 m/s [from conservation of energy]
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b. N top = m ( v 2 /R – g) = 21x10 3 N 5) Energy and Friction a. v bottom = √( 2gh) = 7.67 m/s b. E friction = mgh = 58.8 J [since the block comes to rest, all energy goes into friction] c. μ k = h/d = 1/3 [using E friction = ƒ k d, the work done by friction]
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Practice_Final_Solutions - Solutions to Phys. 6A Practice...

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