As can be seen from the figure, when the elevator accelerates, the pulling force on the heavy object is F 1=58N, and the resultant force on the heavy object is F 1 = F 1-mg. According to Newton's second law, some weights will produce acceleration.
A 1 = f or 1m = 58? 5×105m/s2 =1.6m/s2
When the elevator decelerates, the pulling force F2 on the weight is 46n, the resultant force F2 on the weight is F2-mg, and the acceleration generated by the weight according to Newton's second law.
A2 = f = 2m = 46？ 5×105m/s2 =-0.8m/s2
The negative sign indicates that the direction of acceleration is opposite to the direction of tension.
(2) The object moves at a uniform speed after 3 seconds of uniform acceleration, and the speed of uniform movement is the final speed of uniform acceleration, so
The velocity at the end of 3s is v = a1t1=1.6× 3m/s = 4.8m/s.
(3) When the elevator is accelerating uniformly, the rising distance h1=12a1t 21=1.6× 32m = 7.2m..
The distance of the elevator rising at a constant speed h2=vt2=4.8× 10m=48m.
Elevator deceleration rising distance H3 = vt+12a2t23 = 4.8× 6+12× (? 0.8) × 62m =14.4m.
Therefore, the lift height within 19s is h = h1+H2+H3 = 69.6m.
Answer: (1) The initial acceleration of the elevator is a 1= 1.6m/s2, and the final deceleration is 0.8m/S2；; ;
(2) the elevator speed v is 4.8m/s during 3.0 ~ 13.0s;
(3) The rising height of the elevator within 19.0s is h = 69.6m 。