Question

The coefficient of static friction, µs, between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move ? The string and the pulley are assumed to be smooth and massless.

• A. 4.0 kg
• B. 0.2 kg
• C. 0.4 kg
• D. 2.0 kg

Answer 1

Answer: (C)

0.4 kg

Answer 2

The maximum mass value of block B to prevent the two blocks from moving is determined by the maximum static friction force (Fmax). This force can be calculated using the coefficient of static friction (µs) and the weight of block A (WA).

Given:

• Mass of block A (mA) = 2 kg
• Coefficient of static friction (µs) = 0.2

First, calculate the weight of block A (WA): WA = mA$\times$g WA = 2 kg$\times$9.81 m/s² ≈ 19.62 N

Now, determine the maximum static friction force (F_max): Fmax = µs$\times$N F_max = 0.2$\times$19.62 N ≈ 3.924 N

So, the maximum mass value of block B should not exceed the force of approximately 3.924 N to prevent the two blocks from moving.