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Unformatted text preview: Version 122/ABDCC final 06 Turner (59070) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points It takes 6 J of work to stretch a Hookeslaw spring 6 . 95 cm from its unstressed length. How much the extra work is required to stretch it an additional 15 . 7 cm? 1. 18.7066 2. 27.7603 3. 8.36233 4. 6.84479 5. 57.7262 6. 16.2043 7. 4.97423 8. 12.552 9. 18.4614 10. 17.1564 Correct answer: 57 . 7262 J. Explanation: We begin by determining the value of k from the energy equation solved for k , k = 2 U x 2 = 2(6 J) (0 . 0695 m) 2 = 2484 . 34 J / m 2 We can now determine the energy at the ad ditional displacement, U add = 1 2 k x 2 add = 1 2 2484 . 34 J / m 2 (0 . 0695 m + 0 . 157 m) 2 = 63 . 7262 J The extra work required is just the difference in energy between the two displacements. W = U = U add U = 63 . 7262 J 6 J = 57 . 7262 J 002 10.0 points A 130 kg block is released at height h = 3 . 7 m as shown. The track is frictionless except for a portion of length 6 . 9 m. The block travels down the track, hits a spring of force constant k = 1537 N / m, and compresses it 2 . 2 m from its equilibrium position before coming to rest momentarily. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 7 m 2 . 2 m 6 . 9 m 130 kg 1537 N / m Determine the coefficient of kinetic friction between surface and block over the 6 . 9 m track length. 1. 0.113105 2. 0.527975 3. 0.390973 4. 0.447849 5. 0.202683 6. 0.513716 7. 0.464463 8. 0.15323 9. 0.174255 10. 0.148437 Correct answer: 0 . 113105. Explanation: Let : m = 130 kg , k = 1537 N / m , h = 3 . 7 m , = 6 . 9 m , x = 2 . 2 m , and g = 9 . 8 m / s 2 . From conservation of energy E = W f or k x 2 2 mg h = mg . Hence, = mg h k x 2 2 mg Version 122/ABDCC final 06 Turner (59070) 2 = (3 . 7 m) (1537 N / m) (2 . 2 m) 2 2 (130 kg)(9 . 8 m / s 2 ) (6 . 9 m) = . 113105 . 003 10.0 points An 89 g autographed baseball rolls off of a 1 . 5 m high table and strikes the floor a hori zontal distance of 1 m away from the table. 1 . 5m 1 m How fast was it rolling on the table before it fell off? The acceleration of gravity is 9 . 81 m / s 2 . 1. 1.80831 2. 2.32282 3. 1.96384 4. 1.45683 5. 1.67806 6. 1.26259 7. 1.51614 8. 1.47816 9. 2.05896 10. 1.57338 Correct answer: 1 . 80831 m / s. Explanation: Let : x = 1 m , y = 1 . 5 m , and g = 9 . 81 m / s 2 . v x y x From the vertical motion, since v i,y = 0, y = v i,y t + 1 2 g ( t ) 2 = 1 2 g ( t ) 2 t = radicalBigg 2 y g From the horizontal motion, since a x = 0, x = v i,x t + 1 2 a x ( t ) 2 = v x t v x = x t = radicalbigg g 2 y x = radicalBigg 9 . 81 m / s 2 2( 1 . 5 m) (1 m) = 1 . 80831 m / s ....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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