A force diagram is shown below:
where,
N: normal force on block1
fg1: weight of block1 = (6kg)(9.8m/s^2) = 58.8N
T: tension in the string
Fr: friction force between the block and the table = μk*N
fg2: weight of block2 = (5kg)(9.8m/s^2) = 49N
Consider that the normal force is equal and opposite to the weight of block1 (becasue the equilibrium of the vertical forces n the block1), then:
N = fg1 = 58.8N
μk: coefficient of kinetic friction = 0.3
Fr = μk*N = (0.3)(58.8N) = 17.64N
Based on the given information and the force diagram, you can conclude:
Four forces are acting on the sliding block.
Two forces are acting on the hanging block.
The forces of the system are unbalanced due to the weight of the hanging block.
The magnitud of the friction force is 17.64N
The mass of the system is the sum of the masses of the blocks:
mass of the system = 6kg + 5kg = 11kg
The net force is, based on the clarification of point 4:
Fnet = fg2 - Fr = 49N - 17.64N = 31.36N
The acceleration of the system is:
a = 31.36N/(11kg) = 2.85 m/s^2
After release the sliding block moves to the right.
The tension in hte string is calculated by using the sum of forces on the sliding block:
T = Fr + m1*a
T = 17.64N + (6kg)(2.85m/s^2) = 17.64N + 17.1N = 34.74N