A 10 kg block is connected to an empty 1kg bucket by a cord running over a friction less pulley. The static friction coefficient is 0.4 and the kinetic friction between the table and block is 0.3. Sand is gradually added to the bucket until the system just begins to move. The mass of sand added to the bucket and the acceleration (in m/s2) of the system respectively (g = acceleration due to gravity)

To solve this problem, we need to consider the forces acting on the system and determine the point at which the system just begins to move.

1. First, let's analyze the forces acting on the block and the bucket. The weight of the block is given by W_block = m_block * g, where m_block is the mass of the block (10 kg) and g is the acceleration due to gravity.

2. The weight of the bucket is given by W_bucket = m_bucket * g, where m_bucket is the mass of the bucket (1 kg).

3. Since the block and the bucket are connected by a cord running over a frictionless pulley, the tension in the cord is the same on both sides. Let's denote this tension as T.

4. The force of static friction between the block and the table can be calculated using the equation F_static = μ_static * N, where μ_static is the static friction coefficient (0.4) and N is the normal force. In this case, the normal force is equal to the weight of the block, N = W_block.

5. The force of kinetic friction between the block and the table can be calculated using the equation F_kinetic = μ_kinetic * N, where μ_kinetic is the kinetic friction coefficient (0.3) and N is the normal force. Again, the normal force is equal to the weight of the block, N = W_block.

6. Now, let's consider the forces acting on the system when it is just about to move. At this point, the force of static friction is at its maximum value and is equal to the force applied by the tension in the cord, F_static = T.

7. The force of kinetic friction is not relevant at this point since the system is not yet in motion.

8. Therefore, we can set up the following equation: T = F_static = μ_static * N = μ_static * W_block.

9. Substituting the values, we have T = 0.4 * (10 kg * g).

10. Now, let's calculate the mass of sand added to the bucket. Since the weight of the bucket is equal to the tension in the cord, we can equate it to T.

11. Therefore, m_sand * g = T.

12. Substituting the values, we have m_sand * g = 0.4 * (10 kg * g).

13. Simplifying the equation, we find m_sand = 0.4 * 10 kg = 4 kg.

14. Finally, let's calculate the acceleration of the system. Since the net force acting on the system is equal to the tension in the cord, we can use Newton's second law, F_net = m_total * a, where m_total is the total mass of the system (block + bucket + sand) and a is the acceleration.

15. The net force is given by F_net = T - F_kinetic = T - (μ_kinetic * N) = T - (μ_kinetic * W_block).

16. Substituting the values, we have F_net = T - (0.3 * (10 kg * g)).

17. Since the system is just about to move, the net force is equal to zero, so we can set up the equation T - (0.3 * (10 kg * g)) = 0.

18. Substituting the value of T, we have 0.4 * (10 kg * g) - (0.3 * (10 kg * g)) = 0.

19. Simplifying the equation, we find 0.1 * (10 kg * g) = 0.

20. Therefore, the acceleration of the system is zero, a = 0 m/s^2.

In conclusion, the mass of sand added to the bucket is 4 kg and the acceleration of the system is 0 m/s^2.