Question

A metal block with a base area of $0.20 \ m^2$ rests on a table with a liquid film of thickness 0.25 mm between them. When a horizontal force of 0.1 N is applied, the block moves at a constant speed. If the liquid's viscosity is $5.0 \times 10^{-3}$ Pa-s, what is the speed of the block?

• A. $35 \times 10^{-3}$ m/s
• B. $25 \times 10^{-3}$ m/s
• C. $45 \times 10^{-3}$ m/s
• D. $15 \times 10^{-3}$ m/s

Answer 1

Answer: (B)

25 \times 10^{-3} m/s

Answer 2

The viscous force $F$ is given by $F = \eta A \frac{dv}{dy}$, where $\eta$ is the viscosity, $A$ is the area, and $\frac{dv}{dy}$ is the velocity gradient. For a thin liquid film, $\frac{dv}{dy} \approx \frac{v}{h}$, where $v$ is the block's speed and $h$ is the film thickness. Since the block moves at a constant speed, the applied force equals the viscous force: $F = \eta A \frac{v}{h}$. Rearranging for $v$: $v = \frac{Fh}{\eta A}$. Substituting the given values: $v = \frac{(0.1 \ N)(0.25 \times 10^{-3} \ m)}{(5.0 \times 10^{-3} \ Pa-s)(0.20 \ m^2)} = \frac{0.025 \times 10^{-3}}{1.0 \times 10^{-3}} \ m/s = 0.025 \ m/s = 25 \times 10^{-3} \ m/s$.