Please show steps thank you The system is released from rest

Please show steps. thank you

The system is released from rest. Neglect the friction and the masses of the pulleys. Determine the tension in the cable and the distance traveled by block B at the instant its velocity reaches 2 m/s.

Solution

We have

2aA = 3aB

on B:

mB g (3/5) - 3T = mB aB

24g - 3T = 40 aB

3T + 40aB = 235.2 ......(1)

on A:

2T - 8g = 8 aA

2T - 8g = 8 (3 aB / 2) = 12 aB

2T - 12aB = 78.4. ......(2)

Solve the above two equations.

Hence the tension in the cable is

T = 51.4 N

Acceleration:

aB =2.03 m/s^2

Use the below equation and solve d.

v^2 - u^2 = 2 a d

2^2 - 0 = 2 ( 2.03)d

d = 0.985 m