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,g10m/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】12

【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

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

·············································2

2)物体上升高度········································1

物体上升时间;下落时间得出

总时间:·········································1

落回出发点的速度····························1

因而:·······························2

········································2

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