If the applied force is 60 N and the lower mass is 1.0 kg, what is the tension in the string?(frictionless surface)

The net force on the larger mass in the horizontal direction (where we take the direction of F to be positive) is
F - T = 2Ma
where T is the magnitude of tension and a is the mass's acceleration.
The net force on the smaller mass in the vertical direction (upward is positive) is
T - Mg = Ma
If M = 1.0 kg and F = 60 N, then we end up with two equations with two unknowns,
60 N - T = (2.0 kg) a
T - (1.0 kg) (9.8 m/s²) = (1.0 kg) a
Eliminate a and solve for T :
(60 N - T) - 2 (T - (1.0 kg) (9.8 m/s²)) = (2.0 kg) a - 2 (1.0 kg) a
60 N - T - 2T + 19.6 N = (2.0 kg) a - (2.0 kg) a
40.4 N - 3T = 0
3T = 40.4 N
T = (40.4 N) / 3
T ≈ 13 N