Answered: A 48.0 kg crate of tools rests on a…

Name: A 48.0 kg crate of tools rests on a horizontal floor. You exert a grad
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Description: A 48.0 kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 NN. Then you must reduce your push to 208 NN. to keep it moving at a steady 25

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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A 48.0 kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 NN. Then you must reduce your push to 208 NN. to keep it moving at a steady 25.0 cm/scm/s.

What are the coefficient of static and kinetic friction between the crate and the floor?

Enter your answers numerically separated by a comma.

Expert Solution

Step 1

The crate begins to move when the applied force becomes equal to the maximum static friction between the crate and the floor. It is given by

$\begin{array}{rcl} F_{a} & = & F_{s, m a x} \\ = & μ_{s} N \\ = & μ_{s} m g \end{array}$

Here, $F_{a}$ is the applied force at which the crate starts to move, $F_{s, m a x}$ is the maximum static friction, $μ_{s}$ is the coefficient of static friction, N is the normal force, m is the mass of the crate and g is the acceleration due to gravity.

Rearranging the equation,

$\begin{array}{rcl} μ_{s} & = & \frac{F_{a}}{m g} \\ = & \frac{313 N}{(48.0 kg) (9.80 {m/s}^{2})} \\ = & 0.665 \end{array}$

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