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Homework 35

# Homework 35 - portillo(gdp347 homework 35 Turner(58185 This...

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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 – find 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 Find 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 m v 2 f + 1 2 k x 2 f m v 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 = radicalBig 6 . 89906 m 2 / s 2 = 2 . 62661 m / s . 002 (part 2 of 2) 10.0 points Find the speed of the block after it has moved 0 . 209 m from equilibrium if the coefficient 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 m g x f = 1 2 m v 2 f + 1 2 k x 2 f m v 2 f = 2 F x f - 2 μ k m g x f - k x 2 f v 2 f = 2 F x f - 2 μ k m g 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 = radicalBig 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 floor (as shown in the figure). The 15 kg block is pulled 18 cm down the incline (so that the 25 kg block is 31 cm above the floor) 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 floor (that is, when the spring is unstretched).
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Homework 35 - portillo(gdp347 homework 35 Turner(58185 This...

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