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solutions 9

# solutions 9 - Patel Kinal Homework 9 Due Oct 5 2007 7:00 pm...

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Patel, Kinal – Homework 9 – Due: Oct 5 2007, 7:00 pm – Inst: D Weathers 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A time-varying net force acting on a 1 . 9 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 . 2 m , b = 1 . 3 m / s , d = - 2 . 4 m / s 2 , and e = 1 . 3 m / s 3 , with x in meters and t in seconds. Find the work done on the particle in the first 2 . 9 s of motion. Correct answer: 385 . 227 J. Explanation: Since the force is time dependent W Z x f x i ~ F · d~x = Z x f x i m a dx = m Z x f x i d v dt dx = m Z x f x i d v dx d x dt dx = m Z v f v i v dv = 1 2 m v 2 f - 1 2 m v 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 = d x dt = b + 2 d t + 3 e t 2 v i = 1 . 3 m / s , and (1) v f = (1 . 3 m / s) + 2 ( - 2 . 4 m / s 2 ) (2 . 9 s) +3 (1 . 3 m / s 3 ) (2 . 9 s) 2 = 20 . 179 m / s , (2) where t i = 0 s and t f = 2 . 9 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 (1 . 9 kg) h (20 . 179 m / s) 2 - (1 . 3 m / s) 2 i = (386 . 832 J) - (1 . 6055 J) = 385 . 227 J . keywords: 002 (part 1 of 1) 10 points Robin pushes a wheelbarrow by exerting a 121 N force horizontally. Robin moves it 83 m at a constant speed for 24 s. What power does Robin develop?

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solutions 9 - Patel Kinal Homework 9 Due Oct 5 2007 7:00 pm...

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