**Problem Statement:**A 10.0-kg box is at rest on a concrete floor. The coefficient of static friction between the box and the floor is 0.75, and the coefficient of kinetic friction is 0.62. A person pushes the box by applying a force of 100.0 N. What happens to the box?**Discussion:**To determine whether the box will move, we need to calculate the maximum static friction force that can act on the box. Static friction prevents motion up to a maximum threshold given by the formula:\[ f_s = \mu_s \times N \]where \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force. For a box at rest on a horizontal surface, the normal force \( N \) is equal to the weight of the box:\[ N = m \times g \]where \( m = 10.0 \, \text{kg} \) is the mass of the box, and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.Calculating the normal force:\[ N = 10.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98.0 \, \text{N} \]Now, calculate the maximum static friction force:\[ f_s = 0.75 \times 98.0 \, \text{N} = 73.5 \, \text{N} \]Since the applied force (100.0 N) is greater than the maximum static friction force (73.5 N), the box will overcome static friction and start to move.Once the box is in motion, the kinetic friction force will act on it, which is given by:\[ f_k = \mu_k \times N \]where \( \mu_k \) is the coefficient of kinetic friction.Calculate the kinetic friction force:\[ f_k = 0.62 \times 98.0 \, \text{N} = 60.76 \, \text{N} \]Therefore, once the box starts moving, it will continue to move since the kinetic friction (60.76 N) is less than the applied force (100.0 N).

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Description: **Problem Statement:**A 10.0-kg box is at rest on a concrete floor. The coefficient of static friction between the box and the floor is 0.75, and the coefficient of kinetic friction is 0.62. A person pushes the box by applying a force of 100.0 N. Wh

mass, m = 10 kg static friction coefficient = 0.75 kinetic friction coefficient = 0.62 Force, F =…

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A 10.0-kg box is at rest on a concrete floor. The coefficient of static friction between the box and the floor is 0.75 and the coefficient of kinetic friction is 0.62. A person pushes the box by applying a force of 100.0 N. What happens to the box?

A 10.0-kg box is at rest on a concrete floor. The coefficient of static friction between the box and the floor is 0.75 and the coefficient of kinetic friction is 0.62. A person pushes the box by applying a force of 100.0 N. What happens to the box?

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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