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Q80P

Expert-verifiedFound in: Page 480

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Your thumb squeaks on a plate you have just washed. Your sneakers squeak on the gym floor. Car tires squeal when you start or stop abruptly. You can make a goblet sing by wiping your moistened finger around its rim. When chalk squeaks on a blackboard, you can see that it makes a row of regularly spaced dashes. As these examples suggest, vibration commonly results when friction acts on a moving elastic object. The oscillation is not simple harmonic motion, but is called stick-and-slip. This problem models stick-andslip motion. **

** **

**A block of mass m is attached to a fixed support by a horizontal spring with force constant k and negligible mass (Fig. P15.80). Hooke’s law describes the spring both in extension and in compression. The block sits on a long horizontal board, with which it has coefficient of static friction ${{\mu}}_{{s}}$ and a smaller coefficient of kinetic friction ${{\mu}}_{{k}}$The board moves to the right at constant speed v. Assume the block spends most of its time sticking to the board and moving to the right with it, so the speed v is small in comparison to the average speed the block has as it slips back toward the left. **

** **

**(a) Show that the maximum extension of the spring from its unstressed position is very nearly given by ${{\mu}}_{{s}}$ mg/k. **

(a)Maximum extension of the spring is $\frac{{\mu}_{s}mg}{k}$ .

Kinetic friction =${\mu}_{k}$

Static friction = ${\mu}_{s}$

**If any acting force on any particle at any time is performing the Hooke’s law equation F = -kx**

** then the particle motion will be simple harmonic motion.**

In positive direction of the x the block moves in the board by stretching the spring until the force of the spring *-kx* becomes equal to the maximum force of the static friction in magnitude.

Now by this relation we can get the following equation,

$\begin{array}{c}kx={\mu}_{s}n\\ {\mu}_{s}n={\mu}_{s}mg\\ kx={\mu}_{s}mg\\ x=\frac{{\mu}_{s}mg}{k}\end{array}$

Here *x* is the maximum extension of the spring. The derived equation is equal to the required equation hence it is proved that maximum extension of the spring is $\frac{{\mu}_{s}mg}{k}$

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