牛顿定律应用五瞬时加速度问题.ppt

求瞬时加速度问题求瞬时加速度问题 Evaluation only. Evaluation only. Created with Client Profile . Created with Client Profile . Copyright 2004-2011 Aspose Pty Ltd. Copyright 2004-2011 Aspose Pty Ltd. [思路点拨]解答本类型题时应注意以下两点: (1) 当其他力变化时, 弹簧的弹力不能在瞬间发生变化; (2) 当其他力变化时, 细绳上的拉力可以在瞬间发生变化. Evaluation only. Evaluation only. Created with Client Profile . Created with Client Profile . Copyright 2004-2011 Aspose Pty Ltd. Copyright 2004-2011 Aspose Pty Ltd. AB 如图,两个质量均为 m的重物静止,若剪断绳 OA ,则剪断瞬间 A和B的加速度分别是多少? 0a A =2g a B =0 例题 1: Evaluation only. Evaluation only. Created with Client Profile . Created with Client Profile . Copyright 2004-2011 Aspose Pty Ltd. Copyright 2004-2011 Aspose Pty Ltd. 质量皆为 m的A,B两球之间系着一个不计质量的轻弹簧,放在光滑水平台面上, A球紧靠墙壁,今用力 F将B球向左推压弹簧,平衡后,突然将力 F撤去的瞬间 A,B的加速度分别为多少? ABa A =0 a B =F/m 例题 2: Evaluation only. Evaluation only. Created with Client Profile . Created with Client Profile . Copyright 2004-2011 Aspose Pty Ltd. Copyright 2004-2011 Aspose Pty Ltd. 例3: (2010 ·广东外国语学校模拟)在动摩擦因数μ= 的水平面上有一个质量为 m= 1 kg 的小球,小球与水平轻弹簧及与竖直方向成θ= 45 °角的不可伸长的轻绳一端相连,如图 3-2-,且水平面对小球的弹力恰好为零, 当剪断轻绳的瞬间,取 g= 10 m/s : (1) 此时轻弹簧的弹力大小; (2) 小球的加速度大小和方向; Evaluation only. Evaluation only. Created with Client Profile . Created with Client Profile . Copyright 2004-2011 Aspose Pty Ltd. Copyright 2004-2011 Aspose Pty Ltd. [解题: ] (1) 水平面对小球的弹力为零,小球在绳没有断时受到绳的拉力 F T、重力 mg 和弹簧的弹力 F N作用而处于平衡状态,依据平衡条件得竖直方向有: F T cos θ= mg , 水平方向有: F T sin θ=F N, 解得弹簧的弹力为: F N= mg tan θ= 10 N Evaluation only. Evaluation only. Created with Client Profile . Created with Client Profile . Copyright 2004-2011 Aspose Pty Ltd.