If mA 200 kg and mB 180 kg as shown determine the accelera

If mA = 20.0 kg and mB = 18.0 kg as shown, determine the acceleration of each block. (Start from F = ma for full credit.) If initially mA is at rest 2.00 m from the edge of the table, how long does it take to reach the edge of the table if the system is allowed to move freely? If mB = 1.0 kg, how large must mA be if the acceleration of the system is to be kept at g ? [4.64e+00 m/s2, 9.28e-01 s, 99 kg]

Solution

here,

Case 1 :
mass of block A, ma = 20 kg
mass of block B, mb = 18 kg

From Newton Second law of motion, Sum(F) = 0
for Block B,
F - T = 0
T = mb*g ---------------(1) ( where T is Tension in string)

For Block A
F = ma*a ( Where a is acceleration) ---------------(2)

From 1 and 2,
mB g = (mA + mB)a ...(1)
a = mB g / (mA + mB) ---------------------------(3)
a = (18*9.8)/(20+18)
a = 4.64 m/s^2

Case 2:
Assuming,
s be the distance to the edge of the table,
t be the time taken.
s = at^2 / 2
t = sqrt(2s / a)
t = sqrt(2*2 / 4.64)
t = 0.93 s

Case 3:
Rewriting Eqn 3 for Ma
a = mB g / (mA + mB)
ma + mb = mb*g/a
ma = mb(g/a - 1)
ma = 1(100- 1)
ma = 99 kg