- #1

kostoglotov

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## Homework Statement

A tradesman sharpens a knife by pushing it against a grindstone with a given applied force. The grindstone is 28kg, 15 cm in radius, is originally going at 200 rpm but is slowed down to 180 rpm over 10 s by applying the knife. The coefficient of kinetic friction is 0.20. What is the applied force.

Text answer: 2.2 N

My answer: 0.022 N

## Homework Equations

[tex]\alpha = \frac{\Delta w}{\Delta t}[/tex]

[tex]\tau = \alpha I[/tex]

[tex]F_{net} = \frac{\tau}{r} = 0.20 F_{app}[/tex]

[tex]I = \frac{1}{2}MR^2[/tex]

[tex]\frac{rad}{s} = rpm\frac{2\pi}{60}[/tex]

## The Attempt at a Solution

[tex]\alpha = \frac{-20}{10} = -2 \frac{rpm}{s}[/tex]

[tex]-2 \frac{rpm}{s} = \frac{-\pi}{15}\frac{rad}{s^2}[/tex]

[tex]I = \frac{1}{2}28(0.15)^2 = 3.15\times 10^{-3} kg.m^2[/tex]

[tex]\tau = \left|\frac{-\pi}{15}\right|3.15\times 10^{-3} \approx 6.6\times 10^{-3} N.m[/tex]

[tex]F_{net} = \frac{6.6\times 10^{-3}}{0.15} = 4.4\times 10^{-3} N[/tex]

[tex]F_{app} = \frac{4.4\times 10^{-3}}{0.20} = 0.022 N[/tex]