PQ

ByOm Sharma

According to the question,

    Applying Newton’s II law m g \sin 30-\mu_{k} N=m a
    N=m g \cos 30^{\circ}
    m g \sin 30^{\circ}-\mu_{k} m g \cos 30^{\circ}=m a
    \frac{g}{2}-\frac{\sqrt{3}}{2}\left(\mu_{k} g\right)=\frac{g}{4} \quad\left{\right. given \left.a=\frac{g}{4}\right}
    \mu_{k}=\frac{1}{2 \sqrt{3}}

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