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2019-2020学年人教版八年级物理下册期末复习精练:11.1功(原卷版和解析版)

ID:133041 2020年06月23日 资料纠错 有奖上传 收藏
【内容简介】11.1           功1、功的初步概念:如果一个力作用在物体上,物体在这个力的方向上移动了一段距离,就说这个力做了功。2、功包含的两个必要因素:一是作用在物体上的力,二是物体在这个力的方向上移动的距离。3、功的计算:功等于力与物体在力的方向上通过的距离的乘积(功=力×力的方向上的距离)。4、功的计算公式:W=Fs用F表示力,单位是牛(N),用s表示距离,单位是米(m),功的符号是W,单位是牛•米,它有一个专门的名称叫焦耳,焦耳的符号是J,1 J=1 N•m。5、在竖直提升物体克服重力做功或重力做功时,计算公式可以写成W=Gh;在克服摩擦做功时,计算公式可以写成W=fs。6、功的原理;使用机械时,人们所做的功,都不会少于不用机械时(而直接用手)所做的功,也就是说使用任何机械都不省功。7、当不考虑摩擦、机械自身重等因素时,人们利用机械所做的功(Fs)等于直接用手所做的功(Gh),这是一种理想情况,也是最简单的情况。三、知识点精炼一.选择题(共8小题)1.(2020•江西模拟)下列生活实例中,对图片描述正确的有(  ) A.甲图:不计阻力及能量损耗,网球从刚击球拍到球拍形变最大过程中,网球机械能守恒B.乙图:铁锁来回摆动最终停下,在铁锁下降过程中,重力势能全部转化为动能C.丙图:人造地球卫星由于不受空气阻力,只有动能和势能的转化D.丁图:运动员从高处落下,动能转化为重力势能【考点】:力是否做功的判断. 【答案】:C【解析】:A、不计阻力及能量损耗,网球与球拍的总能量是守恒的;但网球从刚撞击球拍到球拍形变最大过程中,网球的机械能会转化为球拍的弹性势能,所以网球的机械能减少,故A错误;B、铁锁来回摆动最终停下;在铁锁下降过程中,由于要克服空气阻力做功,所以重力势能没有全部转化为动能,故B错误;C、在大气层外,卫星绕地球运动时,没有阻力作用,没有克服摩擦做功,只有动能和势能的转化,故C正确;D、运动员从高处落下,质量不变,高度减小,速度增大,则重力势能减小,动能增大,所以重力势能转化为动能,故D错误。故选:C。2.(2020•朝阳区模拟)如图为一些生活中常见的情景,对于这些情景,下列说法中正确的是(  )A. 人用力推箱子但没有推动,人对箱子的推力做了功B. 起重机吊着货箱匀速竖直上升的过程中,起重机对货箱的拉力做了功C. 小学生沿平直道路骑车行进的过程中,地面对车的支持力做了功D. 踢出去的足球在空中飞行的过程中,人脚对足球的作用力做了功【考点】:力是否做功的判断. 【答案】:B【解析】:A、用力推桌子但没有推动,对桌子有力的作用,但桌子没有在力的方向上移动距离,所以不做功;故A错误;B、起重机在提升货箱的过程中,起重机给货箱一个向上的力,重物在力的方向上通过了距离,所以做了功,故B正确;C、平直道路骑车行进的过程中,地面对车的支持力与距离的方向互相垂直,所以不做功;故C错误;D、踢出去的足球继续向前运动的过程中,足球由于惯性向前运动,不再受到脚的作用力,所以脚对球的力对球没有做功。故D错误。故选:B。3.(2020•福建一模)如下图的三种场景中,拉力F1、F2、F3大小相等,在拉力的作用下物体移动的距离相等。若拉力所做的功分别记为W1、W2、W3,下列关于它们大小关系的判断中正确的是(  ) A. W1=W2<W3                 B.W1=W2=W3                 C.W2<W1=W3D.W1<W2<W3【考点】:功的大小比较. 【答案】:A【解析】:由图可见,第一幅图中滑轮为定滑轮,它的特点是不省力,不省距离,所以拉力移动的距离等于物体移动距离为s,拉力做的功为:W1=F1s;第二幅图中,拉力移动的距离也等于物体移动距离为s,拉力做的功为:W2=F2s;第三幅图中滑轮为动滑轮,可以省一半力,但是费距离,拉力移动距离为2s,拉力做的功为:W3=F3×2s。又因为拉力F1、F2、F3大小相等,所以综上分析可得:W1=W2<W3.故A正确,BCD错误。故选:A。
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    所属专辑: 2019-2020学年人教版八年级物理下册期末复习精练

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    2019-2020学年人教版八年级物理下册期末复习精练:11.1功(原卷版和解析版)