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  4. A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of 50 N. The coefficient of friction between the surfaces of contact is ( g =10 m / s 2)

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Q. A block of mass $15\, kg$ is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of $50\, N$. The coefficient of friction between the surfaces of contact is $\left( g =10\, m / s ^{2}\right)$Physics Question Image

Solution:

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$T =50\, N$
Component of force (in y direction)
$N=\frac{T}{\sqrt{2}}+\frac{150}{\sqrt{2}}$
$\Rightarrow N=200 / \sqrt{2}$
$f=\mu \frac{200}{\sqrt{2}}$
Component of force in $x$ direction
$\frac{150}{\sqrt{2}}=\frac{ T }{\sqrt{2}}+ f _{ r }$
$\frac{150}{\sqrt{2}}-\frac{50}{\sqrt{2}}=\frac{\mu \times 200}{\sqrt{2}}$
$\Rightarrow \mu=\frac{1}{2}$