HW04-solutions - markowitz(am45362 HW04 distler(56295 This...

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markowitz (am45362) – HW04 – distler – (56295) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points 1370 N 255 kg μ = 0 . 35 A crate is pushed horizontally with a force. The acceleration of gravity is 9 . 8 m / s 2 . Calculate the acceleration of the crate. Correct answer: 1 . 94255 m / s 2 . Explanation: Let : m = 255 kg , F = 1370 N , and μ = 0 . 35 . 1370 N μ m g a m g The force due to friction is F f = μ m g . Friction acts against the motion, so F net = m a net = F - μ m g a net = F - μ m g m = 1370 N - 874 . 65 N 255 kg = 1 . 94255 m / s 2 , where μ m g = (0 . 35) (255 kg) (9 . 8 m / s 2 ) = 874 . 65 N . 002(part1of3)10.0points A 3.67 kg ball is dropped from the roof of a building 162.7 m high. While the ball is falling to Earth, a horizontal wind exerts a constant force of 12.8 N on the ball. How long does it take to hit the ground? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 5 . 75936 s. Explanation: Let : m = 3 . 67 kg , Δ y = - 162 . 7 m , and g = 9 . 81 m / s 2 . Since v i,y = 0 m/s, Δ y = - 1 2 g t ) 2 Δ t = radicalBigg 2 Δ y - g = radicalBigg 2 ( - 162 . 7 m) - 9 . 81 m / s 2 = 5 . 75936 s . 003(part2of3)10.0points How far from the building does the ball hit the ground? Correct answer: 57 . 8446 m. Explanation: Let : F wind = 12 . 8 N . F x = m a x a x = F wind m = 12 . 8 N 3 . 67 kg = 3 . 48774 m / s 2 . Since v i,x = 0 m/s, Δ x = 1 2 a x t ) 2 = 1 2 ( 3 . 48774 m / s 2 ) (5 . 75936 s) 2 = 57 . 8446 m .
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markowitz (am45362) – HW04 – distler – (56295) 2 004(part3of3)10.0points What is its speed when it hits the ground? Correct answer: 59 . 9639 m / s. Explanation: v x = a x Δ t = ( 3 . 48774 m / s 2 ) (5 . 75936 s) = 20 . 0871 m / s and v y = - g Δ t = - ( 9 . 81 m / s 2 ) (5 . 75936 s) = - 56 . 4993 m / s , so v = radicalBig v 2 x + v 2 y = radicalBig (20 . 0871 m / s) 2 + ( - 56 . 4993 m / s) 2 = 59 . 9639 m / s . 005(part1of2)10.0points A student decides to move a box of books into her dormitory room by pulling on a rope at- tached to the box. She pulls with a force of 170 N at an angle of 27 above the horizon- tal. The box has a mass of 26 . 4 kg, and the coefficient of friction between box and floor is 0 . 256. The acceleration of gravity is 9 . 8 m / s 2 . 26 . 4 kg μ = 0 . 256 170 N 27 Find the acceleration of the box. Correct answer: 3 . 97714 m / s 2 . Explanation: Given : F = 170 N , m = 26 . 4 kg , α = 27 , μ = 0 . 256 , and g = 9 . 8 m / s 2 . F α m g N f k Perpendicular to the floor, F y,net = summationdisplay F y = N + F y - W = 0 N = W - F y = mg - F sin α = (26 . 4 kg) ( 9 . 8 m / s 2 ) - (170 N) sin 27 = 181 . 542 N Parallel to the floor, F x,net = summationdisplay F x = F x - f k = m a x , where f k = μ n , so the horizontal acceleration is a x = F x - μ n m = F cos α - μ n m = (170 N) cos 27 - 0 . 256 (181 . 542 N) 26 . 4 kg = 3 . 97714 m / s 2 . along the floor. 006(part2of2)10.0points The student now starts moving the box up a 10 incline, keeping her 170 N force directed at 27 above the line of the incline. 26 . 4 kg 170 N 27 10
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markowitz (am45362) – HW04 – distler – (56295) 3 If the coefficient of friction is unchanged, what is the new acceleration of the box? Correct answer: 2 . 3135 m / s 2 . Explanation: Given : θ = 10 .
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