# Hw18 - hinojosa(jlh3938 – homework 18 – Turner –(58185 1 This print-out should have 12 questions Multiple-choice questions may continue on

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Unformatted text preview: hinojosa (jlh3938) – homework 18 – Turner – (58185) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A block of mass 0 . 5 kg is pushed against a hor- izontal spring of negligible mass, compressing the spring a distance of ∆ x as shown in the fig- ure. The spring constant is 290 N / m. When released, the block travels along a frictionless, horizontal surface to point B , the bottom of a vertical circular track of radius 0 . 6 m, and continues to move up the track. The speed of the block at the bottom of the track is 12 m / s, and the block experiences an aver- age frictional force of 6 N while sliding up the track. The acceleration of gravity is 9 . 8 m / s 2 . m k R v B v T T B ∆ x What is ∆ x ? Correct answer: 0 . 498273 m. Explanation: From conservation of energy, the initial po- tential energy of the spring is equal to the kinetic energy of the block at B . Therefore, we write 1 2 k (∆ x ) 2 = 1 2 mv 2 B ∆ x = radicalBigg mv 2 B k = radicalBigg (0 . 5 kg) (12 m / s) 2 (290 N / m) = . 498273 m . 002 (part 2 of 3) 10.0 points What is the speed of the block at the top of the track? Correct answer: 8 . 67416 m / s. Explanation: The change in the total energy of the block as it moves from B to T is equal to the work done by the frictional force ∆ E = W f E T- E B = W f . The total energy at B is E B = 1 2 mv 2 B = 1 2 (0 . 5 kg) (12 m / s) 2 = 36 J . The work done by the frictional force is W f =- f π R =- (6 N) ( π ) (0 . 6 m) =- 11 . 3097 J . Therefore, the total energy at T is E T = E B + W f = 36 J + (- 11 . 3097 J) = 24 . 6903 J . We can find now the speed of the block at T from 1 2 mv T 2 = E T- mg h T . Since v T 2 = 2 E T m- 2 g h T , = 2 (24 . 6903 J) . 5 kg- 2(9 . 8 m / s 2 ) (1 . 2 m) = 75 . 2411 m 2 / s 2 , then v T = radicalBig 75 . 2411 m 2 / s 2 = 8 . 67416 m / s . hinojosa (jlh3938) – homework 18 – Turner – (58185) 2 003 (part 3 of 3) 10.0 points What is the centripetal acceleration of the block at the top of the track? Correct answer: 125 . 402 m / s 2 . Explanation: The centripetal acceleration at T is a c = v 2 T R = (8 . 67416 m / s) 2 . 6 m = 125 . 402 m / s 2 ....
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## This note was uploaded on 04/08/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Hw18 - hinojosa(jlh3938 – homework 18 – Turner –(58185 1 This print-out should have 12 questions Multiple-choice questions may continue on

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