텍사스4회솔루션

텍사스4회솔루션

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min (mm64262) – Homework 4 – lee – (GEDB00864) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points A frictionless roller coaster is given an ini- tial velocity of v 0 at height h = 22 m . The radius of curvature of the track at point A is R = 25 m. The acceleration of gravity is 9 . 8 m / s 2 . R 2 3 h h h v 0 A B Find the maximum value of v 0 so that the roller coaster stays on the track at A solely because of gravity. Correct answer: 10 . 0631 m / s. Explanation: let : h = 22 m and R = 25 m . At point A , the weight of coaster must be just large enough to supply the centripetal acceleration. Thus m parenleftbigg v 2 A R parenrightbigg = m g , or v 2 A = R g . Applying conservation of mechanical energy from the start to point A , 1 2 m v 2 0 + m g h = 1 2 m v 2 A + m g parenleftbigg 2 h 3 parenrightbigg v 2 0 = v 2 A - 2 g h 3 = R 2 g 2 - 2 g h 3 v 0 = radicalbigg R g - 2 g h 3 = radicalBigg g parenleftbigg R - 2 h 3 parenrightbigg = radicalBigg (9 . 8 m / s 2 ) bracketleftbigg 25 m - 2 (22 m) 3 bracketrightbigg = 10 . 0631 m / s . 002(part2of2)10.0points Using the value of v 0 calculated in question 1, determine the value of h that is necessary if the roller coaster just makes it to point B. Correct answer: 27 . 1667 m. Explanation: If the speed of the coaster is to be zero at point B, conservation of mechanical energy from the start to point B gives 0 + m g h = 1 2 m v 2 0 + m g h = 1 2 m g parenleftbigg R - 2 h 3 parenrightbigg + m g h h = R 2 + 2 h 3 = 25 m 2 + 2 (22 m) 3 = 27 . 1667 m . 003(part1of3)10.0points A block is released from point A on a track ABCD as shown in the figure. Point A is higher than points B , C , and D . The track is frictionless except for a portion BC which has a coefficient of friction μ . The block travels down the track and hits the spring with spring constant k . The acceleration of gravity is 9 . 8 m / s 2 . A 10 kg 4 . 2 m E h μ B C D 3 . 3 m 2 . 8 m k = 142 kN / m F
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min (mm64262) – Homework 4 – lee – (GEDB00864) 2 Note: The distance h is not to scale. If the spring compresses 7 cm, find the coefficient of friction μ . Correct answer: 0 . 19697. Explanation: Let : = 3 . 3 m , m = 10 kg , h = 4 . 2 m , and k = 1 . 42 × 10 5 N / m . BasicConcepts: Conservation of Energy with Dissipative Forces U spring = 1 2 k x 2 . Solution: The initial potential energy of the block is E i = m g h = (10 kg)(9 . 8 m / s 2 )(4 . 2 m) = 411 . 6 J . The friction force on the path where there is friction is F friction = μ m g and so the me- chanical energy lost due to friction is E μ = μ m g ℓ . Thus, at point C the energy of the block is m g h - μ m g ℓ . So, when the block reaches the spring, the total energy is E i - E μ = m g ( h - μ ℓ ) . This must match the total energy stored by the spring when it stops the block E s = E i - E μ 1 2 k x 2 = m g ( h - μ ℓ ) . So μ = 1 bracketleftbigg h - k x 2 2 m g bracketrightbigg (1) = 1 (3 . 3 m) bracketleftbigg 4 . 2 m - (1 . 42 × 10 5 N / m) (0 . 07 m) 2 2 (10 kg) (9 . 8 m / s 2 ) bracketrightbigg = 0 . 19697 .
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