图所示,一个人用与水平方向成= 300角的斜向下的推力F推一个质量为

图所示,一个人用与水平方向成= 300角的斜向下的推力F推一个质量为20 kg的箱子匀速前进,如图(a)所示,箱子与水平地面间的动摩擦因数为=0.40.求:

(1)推力F的大小;

(2)若该人不改变力F的大小,只把力的方向变为与水平方向成300角斜向上去拉这个静止的箱子,如图(b)所示,拉力作用2.0 s后撤去,箱子最多还能运动多长距离?(g取10 m/s2).

答案

13.5 m


解析:

(1)在图(a)情况下,对箱子有

 

由以上三式得F=120 N.

 (2)在图(b)情况下,物体先以加速度a1做匀速运动,然后以加速度a2做匀减速运动直到停止.对物体有

解之得s2=13.5 m.

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