A cubical block of side 'a' and density 'rho' | vedclass

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(A) The block moves with a constant velocity,which implies that the net force acting on the block along the inclined plane is zero.The component of th

Vedclass · Accepted Answer

$\frac{3\rho agt}{5v}$

Vedclass · Answer

(A) The block moves with a constant velocity,which implies that the net force acting on the block along the inclined plane is zero.The component of the gravitational force acting down the plane is $F_g = mg \sin 	heta$.The mass of the cubical block is $m = ho 	imes 	ext{Volume} = ho a^3$.Thus,$F_g = ho a^3 g \sin 37^{\circ}$.Given $\sin 37^{\circ} = \frac{3}{5}$,so $F_g = ho a^3 g \left(\frac{3}{5}ight) = \frac{3}{5} ho a^3 g$.The viscous force $F_v$ acting on the block due to the fluid film is given by Newton's law of viscosity: $F_v = \eta A \frac{dv}{dy}$,where $\eta$ is the coefficient of viscosity,$A = a^2$ is the contact area,and $\frac{dv}{dy} = \frac{v}{t}$.So,$F_v = \eta a^2 \frac{v}{t}$.Equating the forces: $F_v = F_g \implies \eta a^2 \frac{v}{t} = \frac{3}{5} ho a^3 g$.Solving for $\eta$: $\eta = \frac{3 ho a^3 g t}{5 a^2 v} = \frac{3 ho a g t}{5 v}$.

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