1) As long as there is tension in the string connecting the two objects, they will both accelerate together with the same acceleration, 4 m/s². By Newton's second law,
• the net force on the 3-kg object acting parallel to the incline is
∑ F = F - T - f₃ - 3g sin(30°) = (3 kg) (4 m/s²)
• the net force on the 3-kg object perpendicular to the incline is
∑ F = n₃ - 3g cos(30°) = 0
• the net parallel force on the 6-kg object is
∑ F = T - f₆ - 6g sin(30°) = (6 kg) (4 m/s²)
• and the net perpendicular force on the 6-kg object is
∑ F = n₆ - 6g cos(30°) = 0
From the last two equations, it follows that
n₆ = 6g cos(30°) ≈ 50.9223 N
f₆ = 0.2 n₆ ≈ 10.1845 N
T - f₆ - 6g sin(30°) = 24 N ⇒ T ≈ 63.5845 N ≈ 64 N
2) If the objects are pulled up the incline with constant velocity, then the acceleration of either object is zero. The free body diagrams stay the same, so we have the same equations using Newton's second law, the only difference is that each right side is zero.
The new tension in the string is then
T - f₆ - 6g sin(30°) = 0 ⇒ T ≈ 39.5845 N
From the second equation above, we find
n₃ = 3g cos(30°) ≈ 25.4611 N
f₃ = 0.1 n₃ ≈ 2.5461 N
Then in the first equation,
F - T - f₃ - 3g sin(30°) = 0 ⇒ F ≈ 56.8306 N ≈ 57 N