(完整word版)物理学第三版(刘克哲张承琚)课后习题答案第二章.pdf

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1、1 物理学 2 章习题解答 2-1 处于一斜面上的物体，在沿斜面方向的力f 作用下，向上滑动。已知斜面长为5.6 m，顶端的高度为 3.2 m，f 的大小为 100 n，物体的质量为12 kg，物体沿斜面向上滑动的距离为 4.0 m，物体与斜面之间的摩擦系数为0.24。求物体在滑动过程中，力f、摩擦力、重力和斜面对物体支撑力各作了多少功？这些力的合力作了多少功？将这些力所作功的代数和与这些力的合力所作的功进行比较，可以得到什么结论？解 物体受力情形如图2-3 所示。力 f 所作的功；摩擦力,摩擦力所作的功；重力所作的功;支撑力 n 与物体的位移相垂直，不作功，即；这些功的代数和为.物体所受合力

2、为,合力的功为.这表明，物体所受诸力的合力所作的功必定等于各分力所作功的代数和。2-3 物体在一机械手的推动下沿水平地面作匀加速运动，加速度为0.49 ms2。若动力机械的功率有 50%用于克服摩擦力，有 50%用于增加速度，求物体与地面的摩擦系数。解 设机械手的推力为f 沿水平方向，地面对物体的摩擦力为f，在这些力的作用下物体的加速度为 a，根据 牛顿第二定律，在水平方向上可以列出下面的方程式,图 2-3 2 在上式两边同乘以v，得,上式左边第一项是推力的功率()。按题意，推力的功率 p 是摩擦力功率 fv 的二倍，于是有.由上式得,又有,故可解得.2-4 有一斜面长 5.0 m、顶端高 3

3、.0 m，今有一机械手将一个质量为1000 kg的物体以匀速从斜面底部推到顶部，如果机械手推动物体的方向与斜面成30，斜面与物体的摩擦系数为 0.20，求机械手的推力和它对物体所作的功。解 物体受力情况如图2-4 所示。取 x 轴沿斜面向上，y 轴垂直于斜面向上。可以列出下面的方程,(1),(2).(3)根据已知条件,.由式(2)得.将上式代入式(3)，得.将上式代入式(1)得,由此解得图 2-4 文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10

4、V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7

5、E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5

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8、2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编

9、码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A

10、9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H13 .推力 f 所作的功为.2-5 有心力是力的方向指向某固定点(称为力心)、力的大小只决定于受力物体到力心的距离的一种力，万有引力就是一种有心力。现有一物体受到有心力的作用(其中 m 和都是大于零的常量)，从rp 到达 rq，求此有心力所作的功，其中rp和 rq是以力心为坐标原点时物体的位置矢量。解 根据题意，画出物体在有心力场中运动的示意图，即图 2-5，物体

11、在运动过程中的任意点c 处，在有心力 f 的作用下作位移元dl，力所作的元功为,所以，在物体从点p(位置矢量为 rp)到达点 q(位置矢量为 rq)的过程中，f 所作的总功为.2-6 马拉着质量为 100 kg 的雪撬以 2.0 ms1 的匀速率上山，山的坡度为0.05(即每100 m升高 5 m)，雪撬与雪地之间的摩擦系数为 0.10。求马拉雪撬的功率。解 设山坡的倾角为，则.可列出下面的方程式,.式中 m、f、f 和 n 分别是雪橇的质量、马的拉力、地面对雪橇的摩擦力和地面对雪橇的支撑力。从以上方程式可解得,图 2-5 文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T

12、3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H

13、1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:

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16、 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7

17、Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9

18、 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H14 .于是可以求得马拉雪橇的功率为.2-7 机车的功率为 2.0 106 w，在满功率运行的情况下，在100 s内将列车由静止加速到 20 m s1。若忽略摩擦力，试求：(1)列车的质量；(2)列车的速率与时间的关系；(3)机车的拉力与时

19、间的关系；(4)列车所经过的路程。解(1)将牛顿第二定律写为下面的形式,(1)用速度 v 点乘上式两边，得.式中 fv=p，是机车的功率，为一定值。对上式积分,即可得,将已知数据代入上式，可求得列车的质量，为.(2)利用上面所得到的方程式,就可以求得速度与时间的关系，为.(2)(3)由式(2)得,将上式代入式(1)，得文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N1

20、0V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE

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22、5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE

23、10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9

24、S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档

25、编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5

26、A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H15 ,由上式可以得到机车的拉力与时间的关系.(4)列车在这 100 秒内作复杂运动，因为加速度也在随时间变化。列车所经过的路程可以用第一章的位移公式(1-11)来求解。对于直线运动，上式可化为标量式，故有.2-8 质量为 m 的固体球在空气中运动将受到空气对它的黏性阻力f 的作用，黏性阻力的大小与球相对于空气的运动速率成正比，黏性阻力的方向与球的运动方向相反，即可表示为 f=v，其中 是常量。已知球被约束在水平方向上，在空气的黏性阻力作用下作减

27、速运动，初始时刻t0，球的速度为 v0，试求：(1)t 时刻球的运动速度v；(2)在从 t0 到 t 的时间内，黏性阻力所作的功a。解(1)根据已知条件，可以作下面的运算,式中.于是可以得到下面的关系,对上式积分可得.(1)当 t=t0时，v=v0，代入上式可得.将上式代入式(1)，得.(2)(2)在从 t0 到 t 的时间内，黏性阻力所作的功可以由下面的运算中得出文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE

28、10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9

29、S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档

30、编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5

31、A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N1

32、0V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE

33、7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B

34、5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H16 .2-9 一个质量为 30 g 的子弹以 500 m s1 的速率沿水平方向射入沙袋内，并到达深度为 20 cm 处，求沙袋对子弹的平均阻力。解 根据动能定理，平均阻力所作的功应等于子弹动能的增量，即,所以.2-10 以 200 N 的水平推力推一个原来静止的小车，使它沿水平路面行驶了5.0 m。若小车的质量为 100 kg，小车运动时的摩擦系数为0.10，试用牛顿运动定律和动能定理两种方法求

35、小车的末速。解 设水平推力为 f，摩擦力为 f，行驶距离为 s，小车的末速为 v。(1)用牛顿运动定律求小车的末速v：列出下面的方程式,.两式联立求解，解得,将已知数值代入上式，得到小车的末速为.(2)用动能定理求小车的末速v：根据动能定理可以列出下面的方程式,其中摩擦力可以表示为.由以上两式可解得,将已知数值代入上式，得小车的末速为文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:C

36、J5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7

37、N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6

38、HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z

39、7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9

40、ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3

41、W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1

42、文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H17.2-11 质量 m=100 g的小球被系在长度l=50.0 cm绳子的一端，绳子的另一端固定在点 o，如图 2-6 所示。若将小球拉到p 处，绳子正好呈水平状，然后将小球释放。求小球运动到绳子与水平方向成=60 的点 q 时，小球的速率 v、绳子的张力 t 和小球从 p到 q 的过程中重力所作的功a。解 取 q 点的势能为零，则有,即,于是求得小球到达q点时的速率为.设小球到达 q 点时绳子的张力为 t，则沿轨道法向可以列出

43、下面的方程式,由此可解的.在小球从 p 到 q 的过程中的任意一点上，沿轨道切向作位移元ds，重力所作元功可表示为,式中 是沿轨道切向所作位移元ds与竖直方向的夹角。小球从 p 到 q 的过程中重力所作的总功可以由对上式的积分求得.2-12 一辆重量为 19.6 103 n 的汽车，由静止开始向山上行驶，山的坡度为0.20，汽车开出 100 m 后的速率达到 36 km h1，如果摩擦系数为 0.10，求汽车牵引力所作的功。解 设汽车的牵引力为f，沿山坡向上，摩擦力为f，山坡的倾角为。将汽车自身看为一个系统，根据功能原理可以列出下面的方程式,(1),图 2-6 文档编码:CJ5A9X7N10V

44、6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E

45、7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H

46、9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10

47、T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2

48、H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码

49、:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9

50、X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H1文档编码:CJ5A9X7N10V6 HE7E7Z7B5H9 ZE10T3W9S2H18.根据已知条件，可以得出，汽车的质量以及。从方程(1)可以解得.汽车牵引力所作的功为,将数值代入，得.2-13 质量为 1000 kg 的汽车以 36 km h1 的速率匀速行驶，摩擦系数为0.10