A large box of mass M is at rest on a horizontal table whose surface is rough; the coefficient of static friction between the table and the block is μ. The box is connected by a taut string to a small block of unknown mass m, initially held horizontally as shown and then released.

1. Assuming the string remains taut and the large box M does not slip on the table, what is the tension in the string just before block m strikes the table? Answer in terms of M, m, g, and μ (mu)

2. Suppose now that the large box begins to slip just before the block strikes the wall. What is the mass of the block? Answer in terms of M, g, and μ (mu).

To solve these problems, we will use the concept of Newton's second law of motion and apply it to the forces acting on the system.

1. Before block m strikes the table, the tension in the string can be determined by analyzing the forces acting on it. We need to consider both the static friction force acting on the large box M and the tension force in the string.

First, let's consider the forces acting on the large box M:
- Weight (mg) acting downward.
- Normal force (N) acting upward from the table.
- Static friction force (F_friction) acting in the direction opposite to motion (which is zero since the box is at rest).

Since the box is at rest, the sum of the vertical forces must be zero:
N - mg = 0

The tension force in the string is horizontal and can be calculated by considering the horizontal forces:
T - F_friction = 0

The maximum static friction force can be written as: F_friction = μN

Substituting this value back into the equation, we get:
T - μN = 0

Solving for N in the equation N - mg = 0, we obtain N = mg.

Substituting this value into the equation T - μN = 0, we get:
T - μ(mg) = 0

Therefore, the tension in the string just before block m strikes the table is T = μmg.

2. Now, let's consider the scenario where the large box starts to slip just before the block m strikes the wall.

When the box starts to slip, the maximum static friction force is surpassed by the force applied on it due to the tension in the string.

The maximum static friction force can be written as: F_friction = μN

The force applied on the box due to the tension in the string is equal to the tension force: T.

The force applied on the box due to the tension in the string must be greater than the maximum static friction force for the box to start slipping.

Therefore, we can set up the inequality: T > μN

Substituting N = mg into the above inequality, we get:
T > μmg

Since we want to find the mass of the block, we can rearrange the expression: T/μg > m

Therefore, the mass of the block is m > T/μg.

Please note that the inequality m > T/μg represents the minimum mass required for the block m to cause the large box M to slip.