2022年人教版必修一4.3《牛顿第二定律》WORD教案7 .pdf
名师精编优秀教案第 3 单元：牛顿第二运动定律的应用例 1如图 362 所示,质量为 4 kg 的物体静止于水平面上,物体与水平面间的动摩擦因数为0.5,物体受到大小为20,与水平方向成角斜向上的拉力F作用时沿水平面做匀加速运动,求物体的加速度是多大?(取 10 m/s2)解析:以物体为研究对象,其受力情况如图3 63 所示,建立平面直角坐标系把F沿两坐标轴方向分解,则两坐标轴上的合力分,sincosGFFFFFFNyx物体沿水平方向加速运动,设加速度为,则轴方向上的加速度,轴方向上物体没有运动,故y,由牛顿第二定律得0,yyxxmaFmamaF所以0sin,cosGFFmaFFN又有滑动摩擦力NFF物体的加速度a=0.58 m/s2.小结:当物体的受力情况较复杂时,根据物体所受力的具体情况和运动情况建立合适的直角坐标系,利用正交分解法来解.例 2一斜面AB长为 10,倾角为,一质量为 kg 的小物体(大小不计)从斜面顶端A点由静止开始下滑,如图 364所示(取 10 m/s2)(1)若斜面与物体间的动摩擦因数为0.5,求小物体下滑到斜面底端B点时的速度及所用时间.(2)若给小物体一个沿斜面向下的初速度,恰能沿斜面匀速下滑,则小物体与斜面间的动摩擦因数 是多少?图 363 图 364 图 362 名师精编优秀教案解析:(1)以小物体为研究对象,其受力情况如图3 65所 示,建 立 直 角 坐 标 系,把 重 力沿轴 和轴 方 向 分解:sin,cos21mgGmgG小物体沿斜面即轴方向加速运动,设加速度为,则,物体在轴方向没有发生位移,没有加速度则,由牛顿第二定律得,所以又NFF所以22/67.0/)30cos5.030(sin10)cos(sincossinsmsmgmmgmga设小物体下滑到斜面底端时的速度为,所用时间为,小物体由静止开始匀加速下滑,由asvvt2202smsmasv/7.3/1067.022由atvvt0ssavt5.567.07.3(2)小物体沿斜面匀速下滑时,处于平衡状态,其加速度,则在图36的直角坐标中0,0yxaa,由牛顿第二定律,又NFF图 365 yNyxxmaGFFmaFGF12cossinmgFmaFmgN0012yNyxxm aGFFm aFGFcossinmgFmgFN所以文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4名师精编优秀教案所以,58.030tantanNFF小结:若给物体一定的初速度,当 tg 时,物体沿斜面匀速下滑；当tg （mgcosmgsin ）时,物体沿斜面减速下滑；当tg（mgcosmgsin ）时,物体沿斜面加速下滑.例 3静止在水平地面上的物体的质量为2 kg,在水平恒力F推动下开始运动,4 s末它的速度达到4 m/s,此时将F撤去,又经 6 s 物体停下来,如果物体与地面的动摩擦因数不变,求F的大小.解析:物体的整个运动过程分为两段,前 4 s 物体做匀加速运动,后 6 s 物体做匀减速运动.前 4 s2211/1/440smsmtva设摩擦力为F,1maFF后 6 s2222/32/640smsmtva物体所受的摩擦力大小不变,2maF由可求得水平恒力FNNaamF3.3)321(2)(21小结:解决动力学问题时,受力分析是关键,对物体运动情况的分析同样重要,特别是像这类运动过程较复杂的问题,更应注意对运动过程的分析.在分析物体的运动过程时,一定弄清整个运动过程中物体的加速度是否相同,若不同,必须分段处理,加速度改变时的瞬时速度即是前后过程的联系量.分析受力时要注意前后过程中哪些力发生了变化,哪些力没发生变化.例 4 如图 366 所示,质量为的物块A和质量为的物块B与地面的摩擦均不计.在已知水平推力F的作用下,A、B做加速运动.A对B的作用力为多大?解析:取A、B整体为研究对象,其水平方向只受一个力F的作用根据牛顿第二定律知:（）图 366 文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 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ZR4R5A9S10M4名师精编优秀教案取B为研究对象,其水平方向只受A的作用力F1,根据牛顿第二定律知:1故1小结:对连结体(多个相互关联的物体)问题,通常先取整体为研究对象,然后再根据要求的问题取某一个物体为研究对象.文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 ZR4R5A9S10M4文档编码:CD6Z10C9Z6P2 HR10V10B5X2V9 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