A metal block of area 0.10 ; m^2 is connected | vedclass

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(A) The metal block moves to the right due to the tension in the string. Since the block moves at a constant speed,the net force on it is zero. The te

Vedclass · Accepted Answer

$3.46 	imes 10^{-3} \; Pa \; s$

Vedclass · Answer

(A) The metal block moves to the right due to the tension in the string. Since the block moves at a constant speed,the net force on it is zero. The tension $T$ in the string is equal to the weight of the suspended mass $m$.$T = m g = 0.010 \; kg 	imes 9.8 \; m s^{-2} = 0.098 \; N$This tension is balanced by the viscous drag force $F$ acting on the block.$F = \eta A \frac{dv}{dx}$Where $\eta$ is the coefficient of viscosity,$A = 0.10 \; m^2$ is the area,$v = 0.085 \; m s^{-1}$ is the velocity,and $dx = 0.30 \; mm = 0.30 	imes 10^{-3} \; m$ is the film thickness.Rearranging for $\eta$:$\eta = \frac{F \cdot dx}{A \cdot v}$$\eta = \frac{(0.098 \; N) 	imes (0.30 	imes 10^{-3} \; m)}{(0.10 \; m^2) 	imes (0.085 \; m s^{-1})}$$\eta = \frac{0.0294 	imes 10^{-3}}{0.0085} \; Pa \; s$$\eta \approx 3.46 	imes 10^{-3} \; Pa \; s$

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