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A 1.50-kg ball and a 2.00-kg ball are glued together with

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 39E Chapter 14

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 39E

A 1.50-kg ball and a 2.00-kg ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant 165 N/m. and the system is vibrating vertically with amplitude 15.0 cm. The glue connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the ?lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.

Step-by-Step Solution:

Solution 39E Step 1 of 5: (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion At the lowest point the acceleration due to gravity acting on the lighter ball of mass 1.5kg will drag it downward, whereas the heavier ball of mass 2 kg which is directly connected to the spring with force constant k=165 N/m will be acted by the tension force which is opposite to the gravity and therefore the system of balls vibrates with amplitude A=15 cm=0.15 m. Step 2 of 5: (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose. Let L oe the length of the unstretched spring with force constant k=165 N/m. The equilibrium position when both balls were attached, Mass attached, M=2 kg +1.5 kg=3.5 kg force constant k=165 N/m Equilibrium position = original length + extension due to weight X =L +o Using F=W=kx as x=W/k X=L +oW/k) Substituting k=165 N/m , W=Mg=3.5(9.8) X=L +( 3.5(9.).................1 o 165 Step 3 of 5: The new equilibrium position when only heavier is attached Mass, m=2kg 1 X =L +(o/k) Substituting k=165 N/m , W=mg=2(9.8) X =L +o 2(9.)....................2 165

Step 4 of 5

Chapter 14, Problem 39E is Solved
Step 5 of 5

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The full step-by-step solution to problem: 39E from chapter: 14 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This full solution covers the following key subjects: ball, balls, motion, amplitude, point. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. University Physics was written by and is associated to the ISBN: 9780321675460. This textbook survival guide was created for the textbook: University Physics, edition: 13. Since the solution to 39E from 14 chapter was answered, more than 754 students have viewed the full step-by-step answer. The answer to “A 1.50-kg ball and a 2.00-kg ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant 165 N/m. and the system is vibrating vertically with amplitude 15.0 cm. The glue connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the ?lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.” is broken down into a number of easy to follow steps, and 105 words.

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