A solid block of 2.0 kg mass slides steadily at a velocity V along a vertical wall as shown in the figure below. A thin oil film of thickness h = 0.15 mm provides lubrication between the block and the wall. The surface area of the face of the block in contact with the oil film is 0.04 m² . The velocity distribution within the oil film gap is linear as shown in the figure. Take dynamic viscosity of oil as 7×10^-3 Pa-s and acceleration due to gravity as 10 m/s² . Neglect weight of the oil. The terminal velocity V (in m/s) of the block is _________ (correct to one decimal place).

The correct answer is 10.7.

Step-by-step explanation:

First, let us identify the given values from the problem statement and the accompanying schematic diagram:
- Mass of the block, $m=2.0\text{kg}$
- Contact area of the block with the oil film, $A=0.04{\text{m}}^2$
- Thickness of the oil film, $h=0.15\text{mm}=0.15\times {10}^{-3}\text{m}$
- Dynamic viscosity of the oil, $\mu =7\times {10}^{-3}\text{Pa-s}$
- Acceleration due to gravity, $g=10{\text{m/s}}^2$

1. Force Balance on the Block:
As the block slides down the wall at a steady terminal velocity $V$, the net acceleration is zero. This means the downward weight of the block is perfectly balanced by the upward viscous shear force exerted by the oil film:

$F_{\text{gravity}}=F_{\text{viscous}}$

2. Calculation of Weight (Gravity Force):

$F_{\text{gravity}}=m·g=2.0\text{kg}\times 10{\text{m/s}}^2=20\text{N}$

3. Calculation of Viscous Shear Force:
According to Newton's law of viscosity, the shear stress $\tau$ in the oil film (given a linear velocity distribution) is:

$\tau =\mu ·\frac{V}{h}$

The total viscous drag force $F_{\text{viscous}}$ acting on the contact surface area $A$ is:

$F_{\text{viscous}}=\tau ·A=\mu ·\frac{V}{h}·A$

4. Determining the Terminal Velocity:
Equating the gravitational force to the viscous force:

$m·g=\mu ·\frac{V}{h}·A$

Rearranging the equation to solve for the terminal velocity $V$:

$V=\frac{m·g·h}{\mu ·A}$

Substituting the given values:

$V=\frac{2.0\times 10\times 0.15\times {10}^{-3}}{7\times {10}^{-3}\times 0.04}$

Simplify by cancelling out ${10}^{-3}$ in the numerator and denominator:

$V=\frac{20\times 0.15}{7\times 0.04}$

$V=\frac{3}{0.28}$

$V=\frac{300}{28}=10.714\text{m/s}$

Rounding to one decimal place, we obtain:
$V\approx 10.7\text{m/s}$