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题目.电动机通过一绳子吊起质量为8kg的物体时,绳的拉力不能超过120N,电动机的功率不能超过1200W,要将此物体由静止起用最快的方式吊高60m(经分析,此物体在被吊高接近60m时恰好以最大速度匀速上升)所需时间为多少?(g取10m/s2)这道题在目前的多种高中物理教辅资料的“机械能”一章中较常见,提供的解析基本一致,现摘录如下.原解析:物体要想最快被吊高至60m处,显然要使用最大值条件——即绳的拉力为120N,电动机的功率为1200W,但两个最大值条件不能同时使用.因为刚开始时物体速度很小,若按最大功率运行绳的拉力必然远超出绳
Title: When a motor hoists a mass of 8kg through a rope, the rope’s tension must not exceed 120N. The motor’s power must not exceed 1200W. This object must be lifted by 60m in the fastest way from a standstill (after analysis, this object is What is the time required to be lifted evenly at the maximum speed when the hoisted height is near 60m?) (g takes 10m/s2) This problem is more common in the “mechanical energy” chapter of various current high school physics teaching materials. The analysis provided is basically the same, now extracted as follows. The original analysis: the object to be hung up to 60m at the fastest, apparently to use the maximum condition - that is, the tension of the rope is 120N, the power of the motor is 1200W, but two The maximum condition cannot be used at the same time. Because the speed of the object is very small at the beginning, if the rope is pulled at the maximum power, the pulling force must be far beyond the rope.