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solns2B

# solns2B - ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 k mg k ⎛ ⎝ ⎜...

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EXAM 2B – SOLUTIONS Page 1 of 1 Part II. Word Problems [30 pts each] 1) (A) Use conservation of total mechanical energy: 1 2 mv 2 = mgh h = v 2 2 g = 3.26 m = L (1 cos θ ) θ = 54.4 ° (B) At the bottom of the rope, F Net = T mg = mv 2 L . T = m ( g + v 2 / L ) = 901 N 2) Do (B) first! Energy conservation: E initial + W fric = E final mg ( d + x max ) fd = 1 2 kx max 2 Use quadratic formula: x max = 1.03 m (A) The elevator reaches maximum speed at a height y where force of gravity = spring force mg = k ( x max y ) or y = x max mg k Energy conservation again: E initial + W fric = E final mg ( d + x max ) fd = 1 2 mv max 2 + mg x max mg k
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Unformatted text preview: ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + 1 2 k mg k ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 v max = 7.78 m/s 3) Total momentum conservation: P x before = P x after (200)(5.0 × 10 6 cos30 ° ) = (80)(1.4 × 10 6 cos50 ° ) + 120 v x v x = 6.62 × 10 6 m/s P y before = P y after (200)(5.0 × 10 6 sin30 ° ) = (80)(1.4 × 10 6 sin50 ° ) + 120 v y v y = 3.45 × 10 6 m/s (A) v = v x 2 + v y 2 = 7.46 × 10 6 m/s (B) = tan − 1 ( v y / v x ) = 27.5 ° L L h — d + x max — x max...
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