A block of mass m = 2.00 kg is sitting stationary on a table. A horizontal force ofmagnitude F= 5.00 N is applied to the block, as shown. The coefficient of staticfriction is us = 0.700, and the coefficient of kinetic friction is PK = 0.500. What is the%3Dmagnitude of the force of friction between the block and the table, in units ofnewtons? Enter the numerical answer without units. Your answer must be within 1% ofthe exact answer to receive credit.ls = 0.7HK = 0.5m =2.0 kgF = 5.0 Nblock at rest

Given that Mass m = 2.00 Kg Applied force F = 5.0 N Coefficient of static friction μs = 0.700…

Answered: A block of mass m = 2.00 kg is sitting…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. The block in (Figure 1) has a mass m=10kg and is being pulled by a force F on a table with coefficient of static friction μs=0.3. Four forces act on it: The applied force F (directed θ=30 above the horizontal). The force of gravity Fg=mg (directly down, where g=9.8m/s2). The normal force N (directly up). The force of static friction fsfs (directly left, opposing any potential motion). If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsNfs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: Fcosθ−μsN=0 Fsinθ+N−mg=0 In order to find the magnitude of force FF, we have to solve a system of two equations with both FF and the…

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