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

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

Vedclass · Accepted Answer

$\frac{3\rho a g t}{5v}$

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(A) The block moves with a constant velocity $v$,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 incline is $F_g = mg \sin 	heta$.The viscous drag force acting up the incline is given by Newton's law of viscosity: $F_v = \eta A \frac{v}{t}$,where $A = a^2$ is the contact area.Equating the forces: $F_v = F_g$$\eta (a^2) \frac{v}{t} = mg \sin 37^{\circ}$Given that the mass $m = 	ext{density} 	imes 	ext{volume} = ho a^3$ and $\sin 37^{\circ} = \frac{3}{5}$:$\eta (a^2) \frac{v}{t} = (ho a^3) g \left(\frac{3}{5}ight)$Solving for $\eta$:$\eta = \frac{ho a^3 g \cdot 3}{5 \cdot a^2 \cdot v / t} = \frac{3 ho a g t}{5 v}$

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