why the answer is A and D?why not B?9,如右图所示,小球在竖直力F作用下将竖直弹簧压缩,若将力F撤去,小球将向上弹起并离开弹簧,直到速度变为零为止,在小球上升的过程中 ( )A. 小球的动能先增大后减

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why the answer is A and D?why not B?9,如右图所示,小球在竖直力F作用下将竖直弹簧压缩,若将力F撤去,小球将向上弹起并离开弹簧,直到速度变为零为止,在小球上升的过程中 ( )A. 小球的动能先增大后减
why the answer is A and D?why not B?
9,如右图所示,小球在竖直力F作用下将竖直弹簧压缩,若将力F撤去,小球将向上弹起并离开弹簧,直到速度变为零为止,在小球上升的过程中 ( )
A. 小球的动能先增大后减小
B.小球在离开弹簧时动能最大
c.小球的动能最大时弹性势能为零
D.小球的动能减为零时,重力势能最大
答案是AD

why the answer is A and D?why not B?9,如右图所示,小球在竖直力F作用下将竖直弹簧压缩,若将力F撤去,小球将向上弹起并离开弹簧,直到速度变为零为止,在小球上升的过程中 ( )A. 小球的动能先增大后减
我明白了.
在弹簧恢复原长过程中,弹簧受到阻力源于小球重力,故在恢复原长过程中,弹簧与小球间始终有力的作用.
最开始,弹簧弹力大于小球重力,小球做加速运动.
后来,在一时刻,弹簧弹力与小球重力相等.
之后,弹簧弹力小于小球重力,小球做减速运动,而此时弹性势能不为0.
故无BC

小球离开弹簧 肯定是先加速运动 再减速运动啊 离开弹簧时速度不是最大

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