Under the action of traction F1 and resistance F2 in the horizontal direction, when F1 = F2, F1 is less than F2 and F1 is greater than F2, the car will be ······························································································································?

Under the action of traction F1 and resistance F2 in the horizontal direction, when F1 = F2, F1 is less than F2 and F1 is greater than F2, the car will be ······························································································································?

F 1 = f 2, the car moves in a straight line at a constant speed
If F1 is less than F2, the car will slow down in a straight line
If F1 is greater than F2, the car will accelerate in a straight line

Why is there a backward friction between feet and the car? Why is it not that feet have a forward friction between feet and the car?
Human cart is not a relative backward trend, should it be forward?

According to the law of conservation, the force analysis is as follows
Downward gravity, upward support force:
The forward thrust f requires a backward force friction

As shown in the figure, the car on the smooth table, under the action of the balance force F1 and F2, moves to the right at a uniform speed v. the following statement is wrong
A. If the balance force is reduced to zero at the same time, Untitled will keep the original speed V to move to the right
B. If the force F1 is removed, the object will decelerate and finally come to a standstill
C. If F 2 is removed, the object will accelerate
D. When the car moves to the table, the forces F1 and F2 are removed, and the car will move in a curve after leaving the table
That is to say the reason

I think the mistake should be B, because on a smooth table, the hands are balanced vertically, F1 is removed, or F2 is removed, and the hands are unbalanced horizontally. Therefore, in the end, it is impossible to be stationary, and the motion should be uniformly accelerated

As shown in the figure, the force of the cart used by the person standing on the car is f, and the friction force of the foot to the back of the car is f
A. When the car is moving at a constant speed, the algebraic sum of the work done by F and F to the car is zero. B. when the car is accelerating, the total work done by F and F to the car is negative. C. when the car is decelerating, the total work done by F and F to the car is negative. D. no matter what kind of motion the car is doing, the total work done by F and F to the car is zero

A. When the car is moving in a straight line at a constant speed, the thrust and friction force of the car are equal. The thrust force of the car is doing positive work, and the friction force of the foot is doing negative work. The algebraic sum of the work is 0. So a is correct. B. when the car is accelerating, the friction force of the car is greater than the thrust force of the car, then the positive work of the thrust force of the car is less than that of the foot