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Unformatted text preview: hernandez (ejh742) – homework 31 – Turner – (58120) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 1 . 22 kg block at rest on a tabletop is at tached to a horizontal spring having constant 16 . 9 N / m. The spring is initially unstretched. A constant 25 . 1 N horizontal force is applied to the object causing the spring to stretch. 25 . 1 N 1 . 22 kg 16 . 9 N / m Find the speed of the block after it has moved 0 . 283 m from equilibrium if the sur face between block and tabletop is frictionless. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 3 . 24582 m / s. Explanation: Let : m = 1 . 22 kg , x f = 0 . 283 m , k = 16 . 9 N / m , and F = 25 . 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 (25 . 1 N) (0 . 283 m) 1 . 22 kg (16 . 9 N / m) (0 . 283 m) 2 1 . 22 kg = 10 . 5353 m 2 / s 2 v f = radicalBig 10 . 5353 m 2 / s 2 = 3 . 24582 m / s . 002 (part 2 of 2) 10.0 points Find the speed of the block after it has moved . 283 m from equilibrium if the coefficient of kinetic friction between block and tabletop is . 196. Correct answer: 3 . 07378 m / s. Explanation: Let : μ k = 0 . 196 . 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 (25 . 1 N) (0 . 283 m) 1 . 22 kg 2 (0 . 196) ( 9 . 8 m / s 2 ) (0 . 283 m) (16 . 9 N / m) (0 . 283 m) 2 1 . 22 kg = 9 . 44815 m 2 / s 2 v f = radicalBig 9 . 44815 m 2 / s 2 = 3 . 07378 m / s . 003 10.0 points A 21 kg block is connected to a 35 kg block by a light string that passes over a frictionless pulley having negligible mass. The 40 ◦ incline is smooth (frictionless). The acceleration of gravity is 9 . 8 m / s 2 . The 35 kg block is connected to a light spring of force constant 343 N / m. The spring is unstretched when the 35 kg block is 28 cm above the floor (as shown in the figure). The 21 kg block is pulled 26 cm down the incline (so that the 35 kg block is 54 cm above the floor) and is released from rest. hernandez (ejh742) – homework 31 – Turner – (58120) 2 35 kg T 1 2 1 k g μ = T 2 2 6 c m 40 ◦ 343N / m 28cm 54cm Find the speed of the 21 kg block when the 35 kg block is 28 cm above the floor (that is,...
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Work

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