解析:当物体处于平衡状态时,撤去其中一个力,则其余几个力的合力与撤去的那个力等值反向.题中撤去F1,则物体沿F2的方向由静止开始加速,加速度为a1=;当恢复F1,撤去F2后,由于速度与合力方向相反,所以接下来物体做减速运动,加速度为a2= .由F1=F2,所以a1=a2=0.5 m/s2,由v-t图象:最终物体的速度为零,位移s=×4×1 m=2 m.
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