Ch7-HW1-solutions

# Ch7-HW1-solutions - louey (cal2859) – Ch7-HW1 – florin...

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Unformatted text preview: louey (cal2859) – Ch7-HW1 – florin – (56930) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A horizontal spring with stiffness 0 . 5 N / m has a relaxed length of 13 cm. A mass of 30 g is attached and you stretch the spring to a total length of 29 cm. The mass is then released from rest and moves with little friction. What is the speed of the mass at the moment when the spring returns to its relaxed length of 13 cm? Correct answer: 0 . 653197 m / s. Explanation: From the Conservation of Energy (C.E.) principle, E i = E f , where E i = 1 2 k ( ℓ- ℓ ) 2 , and E f = 1 2 mv 2 f . This implies that v f = ( ℓ- ℓ ) radicalbigg k m = (29 cm- 13 cm) radicalBigg . 5 N / m 30 g = . 653197 m / s . Remember to convert all quantities to ap- propriate units. 002 (part 1 of 5) 10.0 points The figure below is a potential energy curve for the interaction of two neutral atoms. The two-atom system is in a vibrational state in- dicated by the green horizontal line. r 0 eV- 0.2 eV- 0.4 eV- 0.6 eV- 0.8 eV- 1.0 eV- 1.2 eV- 1.4 eV- 1.6 eV r 1 At r = r 1 , what is the approximate value of the kinetic energy K ? Correct answer: 0 . 9 eV. Explanation: The total energy E =- . 4 eV (green horizontal line), and the potential energy U =- 1 . 3 eV (blue curve). Therefore, K = E- U =- . 4- (- 1 . 3) = 0 . 9 eV....
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## This note was uploaded on 03/20/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Ch7-HW1-solutions - louey (cal2859) – Ch7-HW1 – florin...

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