题干

已知某人在地面上最多能举起质量为60kg的物体,而在一个加速下降的电梯里他最多能举起质量为80kg的物体,求此时电梯的加速度大小是多少?(g取10m/s2
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答案(点此获取答案解析)

解:由题意知,此人最大的举力:

F=m1g═60kg×10N/kg=600N,

在加速下降的电梯里,人最多能举起质量为80kg的物体,由牛顿第二定律得:

m2g﹣F=m2a,

解得:a=g﹣ Fm2

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