Explanation Given information: A 33 kg table is pushed across a 6.2 m wide room. The mechanical energy is about 1.5 kJ is converted to internal energy of the table/floor system in the pushing process. Calculation: Both the conservative force and nonconservative force will occur in the table/floor system. Apply the conservation of total mechanical energy equation for nonconservative force system as follows; Δ K + Δ U = W (1) Here, the change in kinetic energy is Δ K , the change in potential energy is Δ U , and the work done by the nonconservative force system is W . Determine the internal energy produced using the relation. W = − f k d = − μ k m g d (2) Here, the coefficient of kinetic friction between table and floor is μ k , acceleration due to gravity is g , and the distance moved is d . Refer to Table 1.1 “SI Prefixes” in the textbook

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ISBN: 9780136780984

Author: Wolfson

Publisher: PEARSON

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Chapter 7, Problem 23E

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To find: the coefficient of kinetic friction between table and floor.

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Chapter 7, Problem 22E

Chapter 7, Problem 24E

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Chapter 7 Solutions

ESSENTIAL UNIV.PHYS.-MOD.MASTERING(18W)

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the coefficient of kinetic friction between table and floor.

Solution Summary: The author calculates the coefficient of kinetic friction between table and floor using the conservation of total mechanical energy equation for nonconservative force system.

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To find: the coefficient of kinetic friction between table and floor.

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Chapter 7 Solutions