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CALC A 3.00-kg fish is attached to the lower end of a

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

Solution for problem 73P Chapter 7

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 73P

CALC A 3.00-kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 900 N/m. The spring initially is neither stretched nor compressed. The fish is released from rest. (a) What is its speed after it has descended 0.0500 m from its initial position? (b) What is the maximum speed of the fish as it descends?

Step-by-Step Solution:

Solution 73P Step 1: Height of the fish from the ground, h = 0.05 m Therefore, the potential energy of the fish due to gravitational force can be calculated. U = mgh Where, m - mass of the fish g - Acceleration due to gravity h - Height of the fish from the ground Provided, mass of the fish, m = 3.00 kg 2 Therefore, U = 3 kg × 9.8 m/s × 0.05 m = 1.47 J Step 2: a) he maximum possible energy of the fish is, U = 1.47 J In the case of spring, the energy at a particular point, E = KE + PE KE of spring = ½ mv 2 PE of spring = ½ kx 2 So, here we can write, ½ mv + ½ kx = 1.47 J (Conservation of energy) We know that, x = 0.05 m and k = 900 N/m 2 2 2 Therefore, ½ {(3 kg × v ) + (900 N/m × 0.05 m )} = 1.47 J 2 2 2 Or, (3 kg × v ) + (900 N/m × 0.05 m ) = 1.47 J × 2 (3 kg × v ) + 2.25 J = 2.94 J 2 (3 kg × v ) = 2.94 J - 2.25 J = 0.69 J v = 0.69 J / 3 kg = 0.23 m /s 2 2 Taking square root on both sides, v = 0.48 m/s The speed of the fish after it descends 0.05 m is, v = 0.48 m/s

Step 3 of 3

Chapter 7, Problem 73P is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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