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1、1 第四章 牛顿运动定律章末总结知识点一、整体法、隔离法分析连接体问题1.连接体两个或两个以上相互作用的物体组成的具有相同加速度的整体叫连接体如几个物体叠放在一起，或并排挤放在一起，或用绳子、细杆等连在一起2处理连接体问题的方法(1)整体法：把整个系统作为一个研究对象来分析的方法不必 考虑系统内力的影响，只考虑系统受到的外力(2)隔离法：把系统中的各个部分(或某一部分)隔离，作为一个单独的研究对象来分析的方法此时系统的内力就有可能成为该研究对象的外力，在分析时要特别注意(3)整体法与隔离法的选用求解各部分加速度都相同的连接体问题时，要优先考虑整体法；如果还需要求物体之间的作用力，再用隔离法求解
2、连接体问题时，随着研究对象的转移，往往两种方法交叉运用一般的思路是先用其中一种方法求加速度，再用另一种方法求物体间的作用力或系统所受合力无论运用整体法还是隔离法，解题的关键还是在于对研究对象进行正确的受力分析【典型例题】【例题 1】如图，两个质量分别为m12 kg、m2=3 kg 的物体置于光滑的水平面上，中间用轻质弹簧秤连接。两个大小分别为F1=30N、F2=20N的水平拉力分别作用在m1、m2上，则A弹簧秤的示数是20 N B弹簧秤的示数是25 N C在突然撤去F2的瞬间，m1的加速度大小为5 m/s2D在突然撤去F1的瞬间，m1的加速度大小为13 m/s2【答案】D【解析】将两物体和弹簧
3、看做一个整体，根据牛顿第二定律可得2512123020/2/5FFamsmsmm，对1m分析可得11FFm a，联立解得11302226FFm aNN，AB错误；在突然撤去2F的瞬间，因为弹簧的弹力不能发生突变，所以1m的受力没有发生变化，故加速度大小仍为22m/s，故 C错误；突然撤去1F的瞬间，1m的受力仅剩弹簧的弹力，对1m列牛顿第二定律得：1Fm a，解得：213/ams，故 D正确2【名师点睛】两个大小分别为123020FNFN、的水平拉力导致物体受力不平衡，先选整体为研究对象进行受力分析，列牛顿第二定律解出加速度，再隔离单独分析一个物体，解出弹簧受力；在突然撤去2F的瞬间，弹簧的弹
4、力不变，对两物块分别列牛顿第二定律，解出其加速度【针 对训练】（多选）如图所示，在光滑的桌面上有M、m的两个物块，现用力F 推物块，使M、m两物块在桌上一起向右加速，则M、m间的相互作用力为A、若桌面光滑，作用力为MFMmB、若桌面光滑，作用力为mFMmC、若桌面的摩擦因数为，M、m仍向右加速，则M、m间的相互作用力为MFMgMmD、若桌面的摩擦因数为，M、m仍向右加速，则M、m间的相互作用力为MFMm【答案】AD【名师点睛】分析整体的受力时采用整体法可以不必分析整体内部的力，分析单个物体的受力时就要用隔离法采用整体隔离法可以较简单的分析问题知识点二、动力学的临界问题1.概念(1)临界问题：某
5、种物理现象(或物理状态)刚好要发生或刚好不发生的转折状态(2)极值问题：在满足一定的条件下，某物理量出现极大值或极小值的情况2关键词语在动力学问题中出现的“最大”、“最小”、“刚好”、“恰能”等词语，一般都暗示了临界状态的出现，隐含了相应的临界条件3常见类型文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编
6、码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D
7、6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y
8、5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K
9、10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8
10、M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5
11、A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z1
12、0C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C23 动力学中的常见临界问题主要有三类：一是弹力发生突变时接触物体间的脱离与不脱离的问题；二是绳子的绷紧与 松弛的问题；三是摩擦力发生突变的滑动与不滑动问题4解题关键解决此类问题的关键是对物体运动情况的正确描述，对临界状态的判断与分析，找出处于临界状态时存在的独特的物理关系，即临界条件常见的三类临界问题的临界条件：(1)相互接触的两个物体将要脱离的临界条件是：相互作用的弹力为零(2)绳子松弛的临界条件是：绳的拉力为零【典型 例题】【例题 2】如图所示，质量为m的光滑小球，用轻绳连接后，挂在三角劈的顶端，绳与
13、斜面平行，劈置于光滑水平面上，斜边与水平面夹角为30，求：(1)劈以加速度a1g/3 水平向左加速运动时，绳的拉力多大？(2)劈的加速度至少多大时小球对劈无压力？加速度方向如何？(3)当劈以加速度a32g向左运动时，绳的拉力多大？【答案】(1)336mg(2)3g，方向水平向左；(3)5mg【解析】(1)如图所示，水平方向：FT1cos FN1sin ma1竖直方向：FT1sin FN1cos mg文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y
14、5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K
15、10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8
16、M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5
17、A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z1
18、0C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编
19、码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D
20、6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C24 由得：FT1336mg.【针对训练】如图所示，有一块木板静止在光滑而且足够长的水平面上，木板的质量为M4 kg、长为L1.4 m，木板右端放着一个小滑块，小滑块质量m1 kg，其尺寸远小于L，小滑块与木板间的动摩擦因数为 0.4.(g取 10 m/s2)(1)现用恒力F作用在木板M上，为使m能从M上面滑落下来，问：F大小的范围是多少？(2)其他条件不变，若恒力
21、F22.8 N，且始终作用在M上，最终使得m能从M上滑落下来，问：m在M上面滑动的时间是多少？【答案】(1)F20 N(2)2 s 知识点三、动力帝的图象问题文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B
22、3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5
23、HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10
24、T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1
25、 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2
26、X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C
27、2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C25 物理图
28、象信息量大，包含知识内容全面，好多习题已知条件是通过物理图象给出的。一、动力学问题中常见的有Ft及aF等图象1at图象，要注意加速度的正负，分析每一段的运动情况，然后结合物体的受力情况根据牛顿第二定律列方程2Ft图象要结合物体受到的力，根据牛顿第二定律求出加速度，分析每一时间段的运动性质3aF图象，首先要根据具体的物理情景，对物体进行受力分析，然后根据牛顿第二定律推导出两个量间的函数关系式，由函数关系式结合图象明确图象的斜率、截距或面积的意义，从而由图象给出的信息求出未知量二、图象在动力学中的应用在物理学问题中，给出已知条件和信息的方式有很多，诸如文字方式、表格方式、函数方式、图象方式，其中图
29、象方式是最常见、最直观的一种 方式，运用图象求解问题也会更加直观、形象1常见的 图象形式在动力学与运动学问题中，常见、常用的图象是位移图象(xt图象)、速度图象(vt图象)和力的图象(Ft图象)等，这些图象反映的是物体的运动规律、受力规律，而绝非代表物体的运动轨迹2图象问题的分析方法遇到带有物理图象的问题时，要认真分析图象，先从它的物理意义、点、线段、斜率、截距、交点、拐点、面积等方面了解图象给出的信息，再利用共点力平衡、牛顿运动定律及运动学公式去解题【典型例题】【例题 3】（多选）如图甲所示，小物块从光滑斜面上由静止开始滑下，斜面保持静止，小物块的位移x和时间的平方t2的关系如图乙所示（g=
30、10m/s2）.下列说法中正确的是A小物块的加速度大小恒为2.5m/s2 B斜面倾角 为30C小物块 2s末的速度是 5m/s D小物块第 2s内的位移为 7.5m【答案】BD【解析】由乙图可知函数关系是：22.5xt，对比位移时间关系212xat可知，物体的加速度为a=5m/s2，选项 A错误；根据a=gsin，可知 sin=0.5，=30，选项B正确；小物块2s 末的速度v=at=10m/s，选项 C错误；小物块第2s 内的位移为221152517.522xmmm，选项 D正确；故选BD.【名师点睛】此题是对牛顿第二定律的应用以及物理图线的考查；关键是能够从物理图线的斜率中得到物体的加速度
31、值，然后根据牛顿第二定律以及运动公式求解各个物理量；对物理图线的考查历来是考试的热文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF
32、5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U
33、1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN
34、7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5
35、C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 Z
36、L5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9
37、Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C26 点问题.【针对训练】质量为 1kg 的物体由静止开始沿光滑斜面下滑，下滑到斜面的底
38、端后进入粗糙水平面滑行，直到静止，它的速度大小随时间的图象如图所示，则斜面的倾角=，物体与水平地面间的动摩擦因数=。【答案】30；0.25【名师点睛】在速度时间图像中，需要掌握三点，一、速度的正负表示运动方向，看运动方向是否发生变化，只要考虑速度的正负是否发生变化，二、图像的斜率表示物体运动的加速度，三、图像与坐标轴围成的面积表示位移，在坐标轴上方表示正方向位移，在坐标轴下方表示负方向位移知识点四、动力学中的传送带问题1摩擦力是否影响传送带的运动是因为带动传送带的电动机在起作用(摩擦力不影响传送带的运动状态)2分析该类问题的关键分析物体与传送带间的滑动摩擦力方向，进而分析物体的运动规律，这是分
39、析传送带问题的关键3常见的传送带模型有两种，一个是水平方向的传送带；另一个是与水平方向成一定角度的传送带(1)物体在水平传送带上的运动有两种可能：若物体到达传送带的另一端时速度还没有达到传送带的速度，则该物体一直做匀变速直线运动；若物体到达传送带的另一端之前速度已经和传送带相同，则物体先做匀变速直线运动后做匀速直线运动(2)对倾斜传送带要分析最大静摩擦力和重力沿斜面方向的分力的关系，如果最大静摩擦力小于重力沿斜面的分力，则物体做匀变速运动；如果最大静摩擦力大于重力沿斜面的分力，则物体做匀速运动【典型例题】【例题 4】水平传送带被广泛地应用于机场和火车站，用于对旅客的行李进行安全检查。如图所示为
40、一水平传送带装置示意图，紧绷的传送带AB始终保持v=1m/s 的恒定速率运行。旅客把行李无初速度地放在A处，文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X
41、9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2
42、文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:C
43、F5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3
44、U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 H
45、N7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T
46、5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C27 设行 李与传送带之间的动摩擦因数=0.1，AB间的距离为
47、2m，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 的恒定速度平行于传送带运动去B处取行李，则A乘客与行李同时到达B B行李提前0.5 s到达 B C乘客提前0.5 s到达 B D若传送带速度足够大，行李会比乘客先到达B【答案】C【名师点睛】关键是判断行李在传送带上的运动性质，需要知道行李无初速度放到传送带上时，在摩擦力的作用下，先做加速运动，这是就有两种可能，一种是加速到和传送带速度相同之前，行李已经到达B点，此时行李做匀加速直线运动；二是在到达B点之前，速度就已经和传送带相同了，此时行李先做匀加速直线运动，后做匀速直线运动【针对训练】（多选）如图所示，绷紧的水平传
48、送带始终以恒定速率v1运行。初速度大小为v2的小物块从与传送带等高的光滑水平地面上的A 处滑上传送带。若从小物块滑上传送带开始计时，小物块在传送带上运动的v-t图象（以地面为参考系）如图乙所示。已知v2 v1，则：At1时刻，小物块离A处的距离达到最大Bt2时刻，小物块相对传送带滑动的距离达到最大C0 t2时间内，小物块受到的摩擦力方向先向右后向左D0 t3时间内，小物块始终受到大小不变的摩擦力作用【答案】AB 文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5
49、D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1
50、Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7K10T5C8M1 ZL5A2X9Z10C2文档编码:CF5D6B3U1Y5 HN7