Explanation 1. Concept: We use kinematic equations to find speed of the refrigerator for certain time and certain distance separately. Also we use angle in the direction of motion and then we can take ratio of speed of refrigerator in case 2 to speed in case 1. 2. Formulae: i. V f = V i + a t ii. V f 2 = V i 2 + 2 a ∆ d iii. F = m a 3. Given: Force applied in horizontal direction and force applied with an angle. 4. Calculations: When refrigerator starts from rest V i = 0 we get. V f = 0 + a t Rearranging for a, a = V f t Using this acceleration in equation of force we get, F x = m a = m V f t F c o s α = m V f t V f = F c o s α t m … … … … . ( 1 ) Similarly we can find a, V f 2 = 0 + 2 a ∆ x a = V f 2 2 ∆ x Using a in force equation, we get F x = m V f 2 2 ∆ x Rearranging it for Vf, V f 2 = F c o s α 2 ∆ x m V f = F c o s α 2 ∆ x m … … … .

Fundamentals of Physics, Volume 1, Chapter 1-20

10th Edition

ISBN: 9781118233764

Author: David Halliday

Publisher: WILEY

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Chapter 5, Problem 84P

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To find:

(a) Ratio of refrigerator’s speed in case 2 to speed in case1 for t.

(b) Ratio of refrigerator’s speed in case 2 to speed in case 1 for d.

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Chapter 5, Problem 83P

Chapter 5, Problem 85P

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Chapter 5 Solutions

Fundamentals of Physics, Volume 1, Chapter 1-20

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(a) Ratio of refrigerator’s speed in case 2 to speed in case1 for t. (b) Ratio of refrigerator’s speed in case 2 to speed in case 1 for d.

Solution Summary: The author explains how they use kinematic equations to find speed of the refrigerator for certain time and certain distance separately.

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To find: (a) Ratio of refrigerator’s speed in case 2 to speed in case1 for t. (b) Ratio of refrigerator’s speed in case 2 to speed in case 1 for d.

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Chapter 5 Solutions

To find:

(a) Ratio of refrigerator’s speed in case 2 to speed in case1 for t.

(b) Ratio of refrigerator’s speed in case 2 to speed in case 1 for d.