Three forces act on an object with a mass of 2kg at the same time, where F1 = 3N, F2 = 4N, F3 = 2n. The directions of F1 and F2 are always vertical, and the direction of F3 can be changed at will, then the acceleration of the object may be () A. 4m/s2B. 3m/s2C. 2m/s2D. 1m/s2

Three forces act on an object with a mass of 2kg at the same time, where F1 = 3N, F2 = 4N, F3 = 2n. The directions of F1 and F2 are always vertical, and the direction of F3 can be changed at will, then the acceleration of the object may be () A. 4m/s2B. 3m/s2C. 2m/s2D. 1m/s2

F1 = 3N, F2 = 4N, the directions of F1 and F2 are always vertical, the resultant force of F1 and F2 is 5N, F3 = 2n, the direction of F3 can be changed at will, the resultant force of three forces: 3N ≤ f ≤ 7n, according to Newton's second law, the acceleration of the object a = FM = F2, 1.5m/s2 ≤ a ≤ 3.5m/s2, so ad is wrong, BC is correct; so select BC

The object with mass m = 5.0kg is placed on a smooth horizontal plane and moves from rest under two mutually perpendicular forces F1 = 3N and F2 = 4N parallel to the horizontal plane
How much work did the two forces do in the first two seconds? What is the average power of force F2 in the second second second? What is the instantaneous power of force F1 at the end of the second second second?

The resultant force is 5 N, and the acceleration is 1
The distance in two seconds is equal to two meters
So the work done by 3N is 3 times 2 times 3 / 5 = 3.6j
The work done by 5N is 4 times 2 times 4 / 5 = 6.4j
In the second second second, the average velocity of the object is 1.5m/s
So the average power is 4 times 1.5 times 4 / 5 = 4.8w
At the end of two seconds, the velocity of the object is equal to 2m / s
The power is 3 times 2 times 3 / 5 = 3.6W

When three forces act on an object with a mass of 2kg at the same time, F1 = 3N, F2 = 4N, F3 = 4N, F1 and F2 remain vertical, F3 direction can be changed arbitrarily, and the acceleration may be

F 1 and F 2 are perpendicular. According to the principle of parallelogram composition of force, the resultant force of F 1 and F 2 is f '= SQR (F 1 ^ 2 + F 2 ^ 2) = 5 (SQR is the root sign). If f 3 and f' are in the same direction, the acceleration is the maximum: Ma = f '+ F 3 = 4 + 5 = 9A = 9 / 2 = 4.5f 3 and f' are opposite, the acceleration is the minimum: Ma = f '+ F 3 = 4-5 = - 1A = - 1 / 2 = - 0.5negative sign is