问题 计算题

“引体向上”是同学们常做的一项健身运动。如图甲所示,质量为m=50kg的某同学两手正握单杠,手臂完全伸直,身体呈自然悬垂状态,此时他的下颚距单杠的高度为H=0.4 m,然后他用F=600 N的恒力将身体向上拉至某位置时不再用力,身体依靠惯性继续向上运动,为保证其下颚超过单杠达到合格要求,如图乙所示。恒力F的作用时间至少为多少?不计空气阻力,取重力加速度g=10 m/s2

答案

t="0.58" s

题目分析:恒力F作用过程

                                                       ①                                                            2分

                                                               ②                                                            2分

                                                                      ③                                                            1分

撤去F

                                                              ④                                                            1分

                                                       ⑤                                                            2分

解得t="0.58" s(或s)                          ⑥                                                             2分

点评:关键是找出第二次引体向上成功的临界条件即末速度等于零.是一道好题.

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