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Homework 10'-solutions

Homework 10'-solutions - kaplan(hmk378 – Homework 10’...

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Unformatted text preview: kaplan (hmk378) – Homework 10’ – Weathers – (17104) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A time-varying net force acting on a 5 . 2 kg particle causes the object to have a displace- ment given by x = a + b t + d t 2 + e t 3 , where a = 2 . 9 m , b = 1 . 1 m / s , d =- 2 m / s 2 , and e = 0 . 5 m / s 3 , with x in meters and t in seconds. Find the work done on the particle in the first 4 . 5 s of motion. Correct answer: 468 . 951 J. Explanation: Since the force is time dependent W ≡ integraldisplay x f x i vector F · dvectorx = integraldisplay x f x i ma dx = m integraldisplay x f x i dv dt dx = m integraldisplay x f x i dv dx dx dt dx = m integraldisplay v f v i v dv = 1 2 mv 2 f- 1 2 mv 2 i . Therefore work done on the particle is the change in kinetic energy. For this case, W = Δ K = K f- K i = 1 2 m ( v 2 f- v 2 i ) , where the velocity is found by differentiating the displacement: v = dx dt = b + 2 dt + 3 et 2 v i = 1 . 1 m / s , and (1) v f = (1 . 1 m / s) + 2 (- 2 m / s 2 ) (4 . 5 s) +3 (0 . 5 m / s 3 ) (4 . 5 s) 2 = 13 . 475 m / s , (2) where t i = 0 s and t f = 4 . 5 s . Evaluation of the velocity at the initial and final times gives the desired result. W = K f- K i = 1 2 m ( v 2 f- v 2 i ) = 1 2 (5 . 2 kg) bracketleftBig (13 . 475 m / s) 2- (1 . 1 m / s) 2 bracketrightBig = (472 . 097 J)- (3 . 146 J) = 468 . 951 J . 002 (part 1 of 3) 10.0 points A 1930 kg car accelerates uniformly from rest to a speed of 10 . 9 m / s in 3 . 74 s....
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Homework 10'-solutions - kaplan(hmk378 – Homework 10’...

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