《第3章 牛顿运动定律高考热点集训(三).pptVIP

速度地放在A处，设行李与传送带之间的动摩擦因数μ＝0.1，A、B间的距离为2 m,g取10 m/s2.若乘客把行李放到传送带的同时也以v＝1 m/s的恒定速率平行于传送带运动 到 B 处取 行 李 , 则(　　) Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. A．乘客与行李同时到达B处 B．乘客提前0.5 s到达B处 C．行李提前0.5 s到达B处 D．若传送带速度足够大，行李最快也要2 s才能到达B处 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 解析：选BD.行李放在传送带上，传送带对行李的滑动摩擦力使行李开始做匀加速直线运动，随后行李又以与传送带相等的速率做匀速直线运动． Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 6．如图所示,薄板A长L＝5 m, 其质 量M＝5 kg,放在水平桌面上,板右端与桌边相齐.在A上距右端s＝3 m处放一物体B(大小可忽略,即可看成质点),其质量m＝2 kg.已知A、B间动摩擦因数μ1＝0.1，A与桌面间和B与桌 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. * 高考热点集训（三） Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 热点1　应用牛顿第二定律解题的常用方法 合成法 若物体只受两个力作用而产生加速度时，利用平行四边形定则求出两个力的合外力方向就是加速度方向． Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 分解法 当物体受到两个以上的力作用而产生加速度时，常用正交分解法，分解方式有两种： (1)分解力：一般将物体受到的各个力沿加速度方向和垂直于加速度方向分 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 解，则：Fx合＝ma(沿加速度方向)，Fy合＝0(垂直于加速度方向)． (2)分解加速度：当物体受到的力相互垂直时，沿这两个相互垂直的方向分解加速度． Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 实战演练 1．如下图所示,在小车中悬挂一小球,若偏角θ未知,而已知摆球的质量为m,小球随小车水平向左运动的加速度为 A ＝ 2g(取g＝10 m/s2) , 则绳的张 力为(　 ) Evaluation only. C