Question

A rectangular block with mass $2\text{kg}$ is sliding down an incline plane at angle of ${30}^{\circ }$. The block is separated from the plane by a lubrication oil with dynamic viscoscity $0.25{\text{Nsm}}^{-2}$ having spacing of $1\text{mm}$. The base of the block has an area of $20\text{cm}\times 50\text{cm}$. Determine the terminal velocity (when acceleration is zero) of the block as it slides down the plane

• A. $39.24\text{cm/s}$
• B. $78.48\text{cm/s}$
• C. $58.86\text{cm/s}$
• D. $19.62\text{cm/s}$

Answer 1

The correct option is A $39.24\text{cm/s}$

Given,
$m=2kg$
$\theta ={30}^0$
$\eta =0.25Ns/m^2$
$t-1\times {10}^{-3}m$
$A=20\times 50\times {10}^{-4}{m}^2$
When all the forces become equal to zero at that time the acceleration will be zero. $i.e.$ $F_{net}=0$
To calculate the net forces on the body the free body diagram can be drawn as

$F_v=mg\sin {30}^0$
$\begin{array}{}{F}_v=\frac{mg}{2}& \text{(1)}\end{array}$

Shear stress, $\tau \propto \frac{dv}{dy}$
$\tau =\eta \frac{dv}{dy}$
$\tau =\eta \frac{v}{y}$
as, $\tau =\frac{F_v}{A}$ replace in equation 1
$\tau \times A=\frac{mg}{2}$
$\eta \frac{v}{y}\times A=\frac{mg}{2}$
$v=\frac{mg}{2}\times \frac{t}{\eta A}$
$v=\frac{2\times 10}{2}\times \frac{1\times {10}^{-3}}{0.25\times 20\times 50\times {10}^{-4}}=\frac{2}{5}=0.4m/s$
$v=40cm/s$