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A 20.0-kg rock is sliding on a rough, horizontal surface

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

Solution for problem 52E Chapter 6

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 52E

A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200. What average power is produced by friction as the rock stops?

Step-by-Step Solution:

Solution 52E Step 1: Consider the data given Mass of rock m = 20 kg Velocity of the down the inclined plane v = 8 m/s Kinetic friction M K = 0.200 Let us find the kinetic energy of the rock This is obtained by using KE = (1/2) mv 2 Substituting the values we get 2 KE = (1/2) × 20 kg × (8.0 m/s) KE = 640 J Thus we find kinetic energy as 640 J Step 2 : We shall find the force acting on the rock This is obtained as F = mg Substituting the values we get F = 20 kg × 9.81 F = 196.2 N Thus we have force on rock as 196.2 N Step 3: Let us find the kinetic friction of the rock K = M × F f K Substituting values we get K = 0.200 × 196.2 N f K = 39.24 N f Thus the kinetic friction of the rock is 39.24 N

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Chapter 6, Problem 52E is Solved
Step 5 of 6

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

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A 20.0-kg rock is sliding on a rough, horizontal surface

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