A metal block of base area 0.20,m^2 is placed | vedclass

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(B) According to Newton's law of viscosity,the viscous force $F$ is given by:$F = \eta A \frac{dv}{dx}$Where:$F = 0.1 \,N$ (applied force)$\eta = 5.0

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$25$

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(B) According to Newton's law of viscosity,the viscous force $F$ is given by:$F = \eta A \frac{dv}{dx}$Where:$F = 0.1 \,N$ (applied force)$\eta = 5.0 	imes 10^{-3} \,Pa-s$ (viscosity)$A = 0.20 \,m^2$ (base area)$dx = 0.25 \,mm = 0.25 	imes 10^{-3} \,m$ (thickness of the film)Since the block moves with a constant speed,the applied force equals the viscous force:$0.1 = (5.0 	imes 10^{-3}) 	imes 0.20 	imes \frac{v}{0.25 	imes 10^{-3}}$$0.1 = \frac{1.0 	imes 10^{-3} 	imes v}{0.25 	imes 10^{-3}}$$0.1 = 4v$$v = \frac{0.1}{4} = 0.025 \,m/s$$v = 25 	imes 10^{-3} \,m/s$Thus,the speed of the block is $25 	imes 10^{-3} \,m/s$.

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