(1) According to the image, the object moves in a straight line at a constant speed twice, and the pulling force is equal to the sliding friction force. Since the pressure and the roughness of the contact surface are the same, the sliding friction force is equal and the pulling force is also equal. That is, F1=F2.
(2) It can be judged from the image that the distance passed by the second time in the same time is small. According to the formula W=FS, when the pulling forces are equal, the greater the distance passed, the work done by the pulling force is The larger it is, that is, W1>W2.
So choose B.