《功能关系能量守恒定律》高三专题复习

功能关系能量守恒

一、功能关系(常见的几种功能对应关系)

重力做正功，重力势能减少；重力做负功，重力势能增加W G＝－ΔE p＝E p1－E p2

弹力做正功，弹性势能减少；弹力做负功，弹性势能增加W弹＝－ΔE p＝E p1－E p2

电场力做正功，电势能减少；电场力做负功，电势能增加W电＝－ΔE p

合外力对物体做功等于物体动能的增量W合＝E k2－E k1

除重力和弹力之外的力做多少正功，物体的机械能就增加多少；除重力和弹力之外的力做多少负功，物体的机械能就减少多少W除G、弹力外＝ΔE

作用于系统的一对滑动摩擦力一定做负功，系统内能增加Q＝F f·x相对

感应电路中产生的焦耳热等于克服安培力做的功,即Q=WA.

二、能量守恒定律

1.内容:能量既不会凭空______，也不会凭空消失，它只能从一种形式______为另一种形式，或者从一个物体______到别的物体，在___________的过程中，能量的总量___________.

2.表达式：ΔE减=______.

针对训练

1如图所示，卷扬机的绳索通过定滑轮用力F拉位于粗糙斜面上的木箱，使之沿斜面加速向上移动．在移动过程中，下列说法正确的是

A．F对木箱做的功等于木箱增加的动能与木箱克服摩擦力所做的功之和

B．F对木箱做的功等于木箱克服摩擦力和克服重力所做的功之和

C．木箱克服重力做的功等于木箱增加的重力势能

D．F对木箱做的功等于木箱增加的机械能与木箱克服摩擦力做的功之和

2．节日燃放礼花弹时，要先将礼花弹放入一个竖直的炮筒中，然后点燃礼花弹的发射部分，通过火药剧烈燃烧产生的高压燃气，将礼花弹由炮筒底部射向空中．若礼花弹在由炮筒底部出发至炮筒口的过程中，克服重力做功W1，克服炮筒阻力及空气阻力做功W2，高压燃气对礼花弹做功W3，则礼花弹在炮筒内运动的过程中(设礼花弹发射过程中质量不变)

A．礼花弹的动能变化量为W3＋W2＋W1

B．礼花弹的动能变化量为W3－W2－W1

C．礼花弹的机械能变化量为W3－W2

D．礼花弹的机械能变化量为W3－W1

3如图所示，固定的倾斜光滑杆上套有一个质量为m的圆环，圆环与竖直放置的轻质弹簧一端相连，弹簧的另一端固定在地面上的A点，弹簧处于原长h.让圆环沿杆滑下，滑到杆的底端时速度为零.则在圆环下滑过程中

A.圆环机械能守恒

B.弹簧的弹性势能先增大后减小

C.弹簧的弹性势能变化了mgh

D.弹簧的弹性势能最大时圆环动能最大

4 在绝缘的平面上方存在着匀强电场方向如图，水平面上的带电金属块在水平拉力F的作用下，沿水平面移动，已知金属块在移动的过程中，外力F做功32J，金属块克服电场力做功8J，金属块克服摩擦力做功16J，则在此过程中金属块的

A．动能增加8J B．电势能增加24J

C．机械能减少24J D．机械能增加48J

5如图所示，实线为电场线，虚线为等势面，相邻两等势面间的电势差相等，一个正电荷在等势面L3处的动能为20J，运动到等势面L1处时动能为零；现取L2为零电势参考平面，则当此电荷的电势能为4J时，它的动能为（不计重力及空气阻力）

A．16J B．10J

C．6J D．4J

6.如图11所示,固定在水平绝缘平面上足够长的金属导轨不计电阻,但表面粗糙,导轨左端连接一个电阻R,质量为m的金属棒(电阻也不计)放在导轨上,并与导轨垂直,整个装置放在匀强磁场中,磁场方向与导轨平面垂直.用水平恒力F把ab棒从静止起向右拉动的过程中,下列说法正确的是

A.恒力F做的功等于电路产生的电能

B.恒力F和摩擦力的合力做的功等于电路中产生的电能

C.克服安培力做的功等于电路中产生的电能

D.恒力F和摩擦力的合力做的功等于电路中产生的电能和棒获得的动能之和

7 如图5所示,将边长为a、质量为m、电阻为R的正方形导线框竖直向上抛出,穿过宽度为b、磁感应强度为B的匀强磁场,磁场的方向垂直纸面向里.线框向上离开磁场时的速度刚好是进入磁场时速度的一半,线框离开磁场后继续上升一段高度,然后落下并匀速进入磁场.整个运动过程中始终存在着大小恒定的空气阻力Ff,且线框不发生转动.求:

(1)线框在下落阶段匀速进入磁场时的速度v2.

(2)线框在上升阶段刚离开磁场时的速度v1.

(3)线框在上升阶段通过磁场过程中产生的焦耳热Q.

8 如图4所示,两条足够长的平行光滑金属导轨,与水平面的夹角均为 ,该空间存在着两个磁感应强度大小均为B的匀强磁场区域Ⅰ和Ⅱ,区域Ⅰ的磁场方向垂直导轨平面向下,区域Ⅱ的磁场方向垂直导轨平面向上,两匀强磁场在斜面上的宽度均为L,一个质量为m、电阻为R、边长为L的正方形金属线框,由静止开始沿导轨下滑,当线圈运动到ab边刚越过ee′即做匀速直线运动;当线框刚好有一半进入磁场区域Ⅱ时,线框又恰好做匀速直线运求:

(1)当线框刚进入磁场区域Ⅰ时的速度v.

(2)当线框刚进入磁场区域Ⅱ时的加速度.

(3)当线框刚进入磁场区域Ⅰ到刚好有一半进入磁场区域Ⅱ的过

程中产生的热量Q.

巩固训练

1.如图所示，在一个盛水的杯子里有一木块.开始时木块被一根细绳拴住而完全没入水中，整个装置与外界绝热，断开细绳，则木块将浮到水面上，最后达到平衡，在这一过程中，水、杯子和木块组成的系统( ) A.内能增大 B.内能减小

C.内能不变

D.条件不足，无法判断

2.滑块静止于光滑水平面上，与之相连的轻质弹簧处于自然伸直状态，现用恒定的水平外力F作用于弹簧右端，在向右移动一段距离的过程中拉力F做了10 J的功.在上述过程中( )

A.弹簧的弹性势能增加了10 J

B.滑块的动能增加了10 J

C.滑块和弹簧组成的系统机械能增加了10 J

D.滑块和弹簧组成的系统机械能守恒

3.(2012·无锡模拟)如图所示，汽车在拱形桥上由A匀速率运动到B，以下说法正确的是( )

A.牵引力与克服摩擦力做的功相等

B.合外力对汽车不做功

C.牵引力和重力做的总功大于克服摩擦力做的功

D.汽车在上拱形桥的过程中克服重力做的功转化为汽车的重力势能

4(2012·盐城调研)如图所示，两物体A、B用轻质弹簧相连，静止在光滑水平面上，现同时对A、B两物体施加等大反向的水平恒力F1、F2，使A、B同时由静止开始运动，在运动过程中，对A、B两物体及弹簧组成的系统，正确的说法是(整个过程中弹簧不超过其弹性限度) ．

A．机械能守恒

B．机械能不断增加

C．当弹簧伸长到最长时，系统的机械能最大

D．当弹簧弹力的大小与F1、F2的大小相等时，A、B两物体速度为零

5如图所示，小球从A点以初速度v0沿粗糙斜面向上运动，到达最高点B后返回A，C为AB的中点.下列说法中正确的是(bd )

A.小球从A出发到返回A的过程中，位移为零，外力做功为零

B.小球从A到C过程与从C到B过程，减少的动能相等

C.小球从A到C过程与从C到B过程，速度的变化量相等

D.小球从A到C过程与从C到B过程，损失的机械能相等

6(2012·常熟模拟)(14分)如图所示,遥控电动赛车(可视为质点)从A点由静止出发,经过时间t后关闭电动机,赛车继续前进至B点后进入固定在竖直平面内的圆形光滑轨道,通过轨道最高点P后又进入水平轨道CD上.已知赛车在水平轨道AB部分和CD部分运动时受到的阻力恒为车重的0.5倍,即k=F f/mg=0.5,赛车的质量m=0.4 kg,通电后赛车的电动机以额定功率P=2 W工作,轨道AB的长度L=2 m,圆形轨道的半径R=0.5 m,空气阻力可以忽略,取重力加速度g=10 m/s2.某次比赛,要求赛车在运动过程中既不能脱离轨道,又要在CD轨道上运动的路程最短.在此条件下,

求:(1)赛车在CD轨道上运动的最短路程.

(2)赛车电动机工作的时间.

7如图所示，ABCD是一个盆式容器，盆内侧壁与盆底BC的连接处都是一段与BC相切的圆弧，B、C在水平线上，其距离d=0.5 m.盆边缘的高度为h=0.30 m.在A处放一个质量为m的小物块并让其由静止出发下滑.已知盆内侧壁是光滑的，而盆底BC面与小物块间的动摩擦因数为μ=0.10.小物块在盆内来回滑动，最后停下来，则停下的位置到B的距离为( )

A.0.50 m

B.0.25 m

C.0.10 m

D.0

8 如图所示，一质量为m的滑块从高为h的光滑圆弧形槽的顶端A处无初速度地滑下，槽的底端B与水平传送带相接，传送带的运行速度恒为v0，两轮轴心间距为l，滑块滑到传送带上后做匀加速运动，滑到传送带右端C时，恰好加速到与传送带的速度相同，求：

(1)滑块到达底端B时的速度大小vB；

(2)滑块与传送带间的动摩擦因数μ；

(3)此过程中，由于克服摩擦力做功而产生的热量Q.

9如图14甲所示,空间存在B=0.5 T,方向竖直向下的匀强磁场,MN、PQ是处于同一水平面内相互平行的粗糙长直导轨,间距L=0.2 m,R是连在导轨一端的电阻,ab是跨接在导轨上质量m=0.1 kg的导体棒.从零时刻开始,通过一小型电动机对ab棒施加一个牵引力F,方向水平向左,使其从静止开始沿导轨做加速运动,此过程中棒始终保持与导轨垂直且接触良好.图乙是棒的v—t图象,其中OA段是直线,AC段是曲线,DE是曲线图象的渐近线,小型电动机在12 s末达到额定功率P额=4.5 W,此后功率保持不变.除R以

外,其余部分的电阻均不计,g=10 m/s2.

(1)求导体棒在0~12 s内的加速度大小.

(2)求导体棒与导轨间的动摩擦因数及电阻R的阻值.（0.2；0.4）

(3)若t=17 s时,导体棒ab达到最大速度,从0~17 s内共发生位移100 m,试求12~17 s内,R上产生的热量是多少?（12.35）

高考英语阅读理解三篇

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