Homework 35 - portillo (gdp347) homework 35 Turner (58185)...

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portillo (gdp347) – homework 35 – Turner – (58185) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A 1 . 47 kg block at rest on a tabletop is at- tached to a horizontal spring having constant 19 . 5 N / m. The spring is initially unstretched. A constant 26 . 3 N horizontal Force is applied to the object causing the spring to stretch. 26 . 3 N 1 . 47 kg 19 . 5 N / m ±ind the speed oF the block aFter it has moved 0 . 209 m From equilibrium iF the sur- Face between block and tabletop is Friction- less. The acceleration oF gravity is 9 . 8 m / s 2 . Correct answer: 2 . 62661 m / s. Explanation: Let : m = 1 . 47 kg , x f = 0 . 209 m , k = 19 . 5 N / m , and F = 26 . 3 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 (26 . 3 N) (0 . 209 m) 1 . 47 kg - (19 . 5 N / m) (0 . 209 m) 2 1 . 47 kg = 6 . 89906 m 2 / s 2 v f = r 6 . 89906 m 2 / s 2 = 2 . 62661 m / s . 002 (part 2 oF 2) 10.0 points ±ind the speed oF the block aFter it has moved 0 . 209 m From equilibrium iF the coe²cient oF kinetic Friction between block and tabletop is 0 . 142. Correct answer: 2 . 51344 m / s. Explanation: Let : μ k = 0 . 142 . Applying the work kinetic energy theorem, F x f - μ k mg x f = 1 2 mv 2 f + 1 2 k x 2 f mv 2 f = 2 F x f - 2 μ k mg x f - k x 2 f v 2 f = 2 F x f - 2 μ k mg x f - k x 2 f m = 2 (26 . 3 N) (0 . 209 m) 1 . 47 kg - 2 (0 . 142) ( 9 . 8 m / s 2 ) (0 . 209 m) - (19 . 5 N / m) (0 . 209 m) 2 1 . 47 kg = 6 . 31737 m 2 / s 2 v f = r 6 . 31737 m 2 / s 2 = 2 . 51344 m / s . 003 10.0 points A 15 kg block is connected to a 25 kg block by a light string that passes over a Frictionless pulley having negligible mass. The 33 incline is smooth (Frictionless). The acceleration oF gravity is 9 . 8 m / s 2 . The 25 kg block is connected to a light spring oF Force constant 248 N / m. The spring is unstretched when the 25 kg block is 13 cm above the ³oor (as shown in the fgure). The 15 kg block is pulled 18 cm down the incline (so that the 25 kg block is 31 cm above the ³oor) and is released From rest.
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portillo (gdp347) – homework 35 – Turner – (58185) 2 25 kg T 1 15 kg μ = 0 T 2 18 cm 33 248 N / m 13 cm 31 cm Find the speed of the 15 kg block when the 25 kg block is 13 cm above the ±oor (that is, when the spring is unstretched).
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This note was uploaded on 01/25/2012 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.

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Homework 35 - portillo (gdp347) homework 35 Turner (58185)...

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