2022八下英语周报第25期答案

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17.【解析】(1)设滑块在斜面上滑行的加速度为a,由牛顿第二定律有mosin 8-prngcos(1分)(1分)联立解得t=2.5s(1分)滑块在圆弧AB上运动过程,由机减能守恒定律有mv+mgR(1-cos6)=÷m(1分其中w=at1(1分)由牛顿第二定律有Fs(1分)联立解得轨道对滑块的支持力FB≈31.7N(1分)(2)滑块在小车上滑行时的加速度a1=p8g=3.5m/s2(1分)小车的加速度a=m=1.5m(1分小车与滑块达到共同速度时小车开始匀速看动满足vB-a1t2=a2t2分N由(1)可知滑块刚滑上小车的速度v=10m/s,最终同速时的速度v=v-a1t2=3m/s(1分)由能量守恒定律可得pmg·n=2mt-2(m+MI(1分)解得5=10m(1分)

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