oldHWK32 - oldhomewk 32 – VALENCIA, DANIEL – Due: Apr...

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Unformatted text preview: oldhomewk 32 – VALENCIA, DANIEL – Due: Apr 28 2008, 4:00 am 1 Question 1, chap 15, sect 3. part 1 of 3 10 points A mass of 378 g connected to a light spring of force constant 32 . 9 N / m oscillates on a horizontal, frictionless track. The amplitude of the motion is 4 . 64 cm. Calculate the total energy of the system. Correct answer: 0 . 0354162 J (tolerance ± 1 %). Explanation: The total energy of the system is equal to its potential energy when the displacement is equal to A : E = 1 2 k A 2 = 1 2 (32 . 9 N / m) (0 . 0464 m) 2 = 0 . 0354162 J . Question 2, chap 15, sect 3. part 2 of 3 10 points What is the maximum speed of the mass? Correct answer: 0 . 432882 m / s (tolerance ± 1 %). Explanation: When the mass is at x = 0, the potential energy is zero and the total energy is E = 1 2 mv 2 max ; therefore, the maximum speed is v max = radicalbigg 2 E m = radicalBigg (2) (0 . 0354162 J) (0 . 378 kg) = 0 . 432882 m / s . Question 3, chap 15, sect 3. part 3 of 3 10 points What is the magnitude of the velocity of the mass when the displacement is equal to . 781 cm? Correct answer: 0 . 426706 m / s (tolerance ± 1 %). Explanation: From conservation of energy, 1 2 mv 2 + 1 2 k x 2 = E = 1 2 k A 2 so the velocity of the mass when the displace- ment is equal to 0 . 781 cm is v = radicalbigg k m ( A 2- x 2 ) = 0 . 426706 m / s . Question 4, chap 15, sect 3. part 1 of 2 10 points A 1 . 26 kg block at rest on a tabletop is at- tached to a horizontal spring having constant 16 . 3 N / m. The spring is initially unstretched. A constant 21 . 1 N horizontal force is applied to the object causing the spring to stretch. The acceleration of gravity is 9 . 8 m / s 2 . 21 . 1 N 1 . 26 kg 16 . 3 N / m Find the speed of the block after it has moved 0 . 277 m from equilibrium if the surface between block and tabletop is frictionless. Correct answer: 2 . 87831 m / s (tolerance ± 1 %). Explanation: Given : m = 1 . 26 kg , x f = 0 . 277 m , k = 16 . 3 N / m , and F = 21 . 1 N . Applying the work kinetic energy theorem, F x f = 1 2 mv 2 f + 1 2 k x 2 f mv 2 f = 2 F x f- k x 2 f v 2 f = 2 F x f- k x 2 f m = 2 (21 . 1 N) (0 . 277 m) 1 . 26 kg oldhomewk 32 – VALENCIA, DANIEL – Due: Apr 28 2008, 4:00 am 2- (16 . 3 N / m) (0 . 277 m) 2 1 . 26 kg = 8 . 2847 m 2 / s 2 so that v f = radicalBig 8 . 2847 m 2 / s 2 = 2 . 87831 m / s ....
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This note was uploaded on 03/26/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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oldHWK32 - oldhomewk 32 – VALENCIA, DANIEL – Due: Apr...

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