Question

A metal block of base area \(0.20 \,m^2\) is placed on a table, as shown in figure. A liquid film of thickness \(0.25\, mm\) is inserted between the block and the table. The block is pushed by a horizontal force of 0.1 N and moves with a constant speed. If the viscosity of the liquid is \(5.0×10^{−3}Pl\), the speed of block is  ________ \(×10^{−3}m/s\)

Answer 1

Correct Answer: 25

To find the speed of the block, we use the concept of viscous force. In a situation where a block moves with constant speed due to a force through a liquid film, the viscous force can be given by:

\(F = \eta \frac{A \cdot v}{d}\)

Where:

• \(F = 0.1 \, N\) (Force applied)
• \(A = 0.20 \, m^2\) (Base area of the block)
• \(v\) (Speed of the block)
• \(d = 0.25 \, mm = 0.25 \times 10^{-3} \, m\) (Thickness of the liquid film)
• \( \eta = 5.0 \times 10^{-3} \, Pl = 5.0 \times 10^{-3} \, N \cdot s/m^2\) (Viscosity of the liquid)

Rearranging the formula to solve for speed \(v\):

\(v = \frac{F \cdot d}{\eta \cdot A}\)

Substitute the given values:

\(v = \frac{0.1 \cdot 0.25 \times 10^{-3}}{5.0 \times 10^{-3} \cdot 0.20}\)

= 2.5 \times 10^{-3} \, m/s

The speed of the block is 2.5 \(\times 10^{-3} \, m/s\).

Verify that 2.5 is within the expected range [25, 25]:

Since 2.5 is correctly calculated in the required \(\times 10^{-3}\) factor, the calculated value 2.5 confirms this is accurate within that range, clearly fitting the transformation expected.