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Unformatted text preview: sanchez (ds28677) – homework 16 – Turner – (58220) 1 This printout should have 12 questions. Multiplechoice 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 205 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 . 5 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 5 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 . 592638 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 (205 N / m) = . 592638 m . 002 (part 2 of 3) 10.0 points What is the speed of the block at the top of the track? Correct answer: 9 . 64282 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 = (5 N) ( π ) (0 . 5 m) = 7 . 85398 J . Therefore, the total energy at T is E T = E B + W f = 36 J + ( 7 . 85398 J) = 28 . 146 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 (28 . 146 J) . 5 kg 2(9 . 8 m / s 2 ) (1 m) = 92 . 9841 m 2 / s 2 , then v T = radicalBig 92 . 9841 m 2 / s 2 = 9 . 64282 m / s . sanchez (ds28677) – homework 16 – Turner – (58220) 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: 185 . 968 m / s 2 . Explanation: The centripetal acceleration at T is a c = v 2 T R = (9 . 64282 m / s) 2 . 5 m = 185 . 968 m / s 2 ....
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This note was uploaded on 04/07/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Mass, Work

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