题目内容
【题目】质量为m=1kg的物体以初速V0=12m/s竖直上抛,空气阻力大小为其重力的0.2倍,g取10m/s2,求:
(1)该物体上升和下降时的加速度之比;
(2)求整个过程中物体克服阻力做功的平均功率P1和物体落回抛出点时重力的瞬时功率P2。
【答案】(1)(2),
【解析】试题(1)受力分析可得:
上升过程中受到向下的重力和空气阻力,牛二定律可得······1分
下降过程中受到向下的重力和向上的空气阻力,牛二定律可得···1分
·············································2分
(2)物体上升高度········································1分
物体上升时间;下落时间得出
总时间:·········································1分
落回出发点的速度····························1分
因而:·······························2分
········································2分
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