HW04-solutions - markowitz(am45362 – HW04 – distler...

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Unformatted text preview: 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 10.0 points 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 μmg a m g The force due to friction is F f = μmg . Friction acts against the motion, so F net = ma net = F- μmg a net = F- μmg m = 1370 N- 874 . 65 N 255 kg = 1 . 94255 m / s 2 , where μmg = (0 . 35) (255 kg) (9 . 8 m / s 2 ) = 874 . 65 N . 002 (part 1 of 3) 10.0 points 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 (Δ t ) 2 Δ t = radicalBigg 2 Δ y- g = radicalBigg 2 (- 162 . 7 m)- 9 . 81 m / s 2 = 5 . 75936 s . 003 (part 2 of 3) 10.0 points 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 = ma 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 (Δ t ) 2 = 1 2 ( 3 . 48774 m / s 2 ) (5 . 75936 s) 2 = 57 . 8446 m . markowitz (am45362) – HW04 – distler – (56295) 2 004 (part 3 of 3) 10.0 points 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 (part 1 of 2) 10.0 points 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 . 256. The acceleration of gravity is 9 . 8 m / s 2 . 26 . 4 kg μ = 0 . 256 1 7 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 mg 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) sin27 ◦ = 181 . 542 N Parallel to the floor, F x,net = summationdisplay F x = F x- f k = ma x , where f k = μn, so the horizontal acceleration is a x = F x- μn m = F cos α- μn m = (170 N) cos27 ◦- . 256(181 . 542 N) 26 . 4 kg = 3 . 97714 m / s 2 .....
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This note was uploaded on 11/29/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas.

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HW04-solutions - markowitz(am45362 – HW04 – distler...

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