Answer to Question #289635 in Mechanics | Relativity for Hin

Question

A block weighing 100 N and having dimensions 200 mm on an edge is

allowed to slide down an incline on a film of oil having a thickness of 0.05 mm.

If a linear velocity profile in the oil is assumed, what is the terminal

speed of the block?

The viscosity of the oil is 7*10^-2 Ns/m^2

Answer 1

Explanation & Calculation

• During the terminal velocity $\small mg\sin \theta=F$. The viscous drag is equal to the component of the weight of the block.
• Moving on, you can use the Newton's law of viscosity for that viscous force to calculate the terminal velocity.

$\qquad\qquad \begin{aligned} \small \frac{F}{A}&=\small\mu.\frac{\Delta v}{\Delta d}\\ \small\frac{mg\sin\theta}{(0.2\,m)^2}&=\small0.07\,Ns^2m^{-2}\times\frac{(V_t-0)}{0.00005\,m} \end{aligned}$

• You know the weight (mg) of the block and the inclination ($\small\theta$) and you can give it a try to find the terminal velocity.