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.

Question

A rocket of initial mass 6000 kg ejects gases at a constant rate of 16 kg/s with a constant relative speed of 11 km/s. What is the acceleration of the rocket one minute after the blast?

Answer 1

To determine the maximum mass of block B such that the blocks remain stationary, we need to understand the forces acting on both blocks A and B.

Block A, which is on the table, experiences static friction resisting the motion caused by the weight of block B. The maximum static friction can be calculated using the formula:

f_{s,\text{max}} = \mu_s \cdot N,

where N is the normal force and \mu_s is the coefficient of static friction.

For block A of mass 2 kg, the normal force N is equal to its weight:

N = m_A \cdot g = 2 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 19.6 \, \text{N}.

Substituting these values into the friction formula gives:

f_{s,\text{max}} = 0.2 \cdot 19.6 = 3.92 \, \text{N}.

For block B to be at rest, the gravitational force (weight) on block B must be less than or equal to the maximum static friction on block A:

m_B \cdot g \leq 3.92 \, \text{N}.

Simplifying the inequality for m_B gives:

m_B \leq \frac{3.92}{9.8} \approx 0.4 \, \text{kg}.

Therefore, the maximum mass of block B that can be supported without causing motion is 0.4 kg.

The correct answer is 0.4 kg.

Answer 2