As shown in the figure, on the smooth slope with an inclination angle of θ = 30 °, there are two wooden blocks a and B connected by light springs. It is known that the mass of a is 2kg, and the mass of B is 3kg. There is a constant force F = 50nd acting on A. at the moment when AB has the same acceleration, remove the external force F. what are the accelerations of a and B at this moment?

As shown in the figure, on the smooth slope with an inclination angle of θ = 30 °, there are two wooden blocks a and B connected by light springs. It is known that the mass of a is 2kg, and the mass of B is 3kg. There is a constant force F = 50nd acting on A. at the moment when AB has the same acceleration, remove the external force F. what are the accelerations of a and B at this moment?

First, the common acceleration before the external force is removed is calculated by the integral method
F-(mA+mB)gsin30°=(mA+mB)a
a+gsin30°=F/(mA+mB)
Taking B as the analysis object, the spring tension is obtained
F1 - mBgsin30° =mBa
F1=mB(a+gsin30°)=FmB/(mA+mB)=50*3/(2+3)=30N
When the external force is removed:
Object a decelerates: F1 + magsin30 ° = maa1
Deceleration: A1 = F1 / MA + gsin30 ° = 30 / 2 + 10 * 1 / 2 = 20m / S ^ 2
Acceleration of object B: F1 + mbgsin30 ° = MBA2
Acceleration: A2 = F1 / mb-gsin30 ° = 30 / 3-10 * 1 / 2 = 5m / S ^ 2