Question

A 196. 2N cubical block, having 20 cm edge, slides down an inclined plane making an angle

of 200 with the horizontal on which there is a thin film of oil having a viscosity of
2. 16*103Ns/m2. What terminal velocity the block will attain, if the film thickness is estimated
to be 0. 025mm?

Answer 1

To determine the terminal velocity of the cubical block sliding down the inclined plane, we need to consider the balance between gravitational force and drag force.

The gravitational force acting on the block can be calculated using the formula:

Gravitational force = mass * gravity

Given that the mass of the block is 196.2 N and the acceleration due to gravity is approximately 9.8 m/s², we have:

Gravitational force = 196.2 N * 9.8 m/s² = 1921.56 N

The drag force acting on the block can be calculated using the formula:

Drag force = viscosity * velocity * surface area

The surface area of the block can be calculated as the square of the edge length:

Surface area = (20 cm)^2 = 400 cm² = 0.04 m²

Given that the viscosity of the oil is 2.16 * 10³ Ns/m² and the thickness of the film is 0.025 mm, we have:

Viscosity = 2.16 * 10³ Ns/m²

Thickness = 0.025 mm = 0.025 * 10⁻³ m

Now we can calculate the velocity at terminal velocity using the formula:

Gravitational force = Drag force

1921.56 N = 2.16 * 10³ Ns/m² * velocity * 0.04 m² / (0.025 * 10⁻³ m)

Simplifying the equation, we find:

velocity = 1921.56 N / (2.16 * 10³ Ns/m² * 0.04 m² / (0.025 * 10⁻³ m))

velocity ≈ 1666.7 m/s

Therefore, the terminal velocity that the cubical block will attain while sliding down the inclined plane is approximately 1666.7 m/s.