This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: hiarker (srh959) – 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 425 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 . 8 m, and continues to move up the track. The speed of the block at the bottom of the track is 14 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 . 480196 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) (14 m / s) 2 (425 N / m) = . 480196 m . 002 (part 2 of 3) 10.0 points What is the speed of the block at the top of the track? Correct answer: 10 . 2138 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) (14 m / s) 2 = 49 J . The work done by the frictional force is W f = f π R = (6 N) ( π ) (0 . 8 m) = 15 . 0796 J . Therefore, the total energy at T is E T = E B + W f = 49 J + ( 15 . 0796 J) = 33 . 9204 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 (33 . 9204 J) . 5 kg 2(9 . 8 m / s 2 ) (1 . 6 m) = 104 . 321 m 2 / s 2 , then v T = radicalBig 104 . 321 m 2 / s 2 = 10 . 2138 m / s . hiarker (srh959) – 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: 130 . 402 m / s 2 . Explanation: The centripetal acceleration at T is a c = v 2 T R = (10 . 2138 m / s) 2 . 8 m = 130 . 402 m / s 2 ....
View
Full
Document
This note was uploaded on 04/23/2010 for the course PHY 303K taught by Professor Nui during the Spring '09 term at University of TexasTyler.
 Spring '09
 NUI
 Physics, Mass, Work

Click to edit the document details