人教版必修二7.7《万有引力定律》同步试题5

中小学教育资源站（)，百万资源免费下载，无须注册！中小学教育资源站版权所有高一物理动能定理机械能守恒检测（计算题）1.“绿色奥运”是2008年北京奥运会的三大理念之一，奥委组决定在各比赛场馆适用新型节能环保电动车，届时奥运会500名志愿者将担任司机，负责接送比赛选手和运输器材。在检测某款电动车性能的某次试验中，质量为8×102kg的电动车由静止开始沿平直公路行驶，达到的最大速度为15m/s,利用传感器测得此过程中不同的时刻电动车的牵引力F与对应的速度v，并描绘出F—1/v图像（图中AB、BO均为直线）。假设电动车在行驶中所受的阻力恒定，求：（1）根据图线ABC，判断该环保电动车做什么运动并计算环保电动车的额定功率（2）此过程中环保电动车做匀加速直线运动的加速度大小（3）环保电动车由静止开始运动，经过多长时间速度达到2m/s?2.如图所示，粗糙的斜面通过一段极小的圆弧与光滑的半圆轨道在B点相连，整个轨道在竖直平面内，且C点的切线水平。现有一个质量为m且可视为质点的小滑块，从斜面上的A点由静止开始下滑，并从半圆轨道的最高点C飞出。已知半圆轨道的半径R=1m,A点到水平底面的高度h=5m,斜面的倾角θ=450，滑块与斜面间的动摩擦因数μ=0.5，空气阻力不计，求小滑块在斜面上的落点离水平面的高度。（g=10m/s2）3.在光滑的水平面有一个静止的物体。现以水平恒力甲推这一物体，作用一段时间后，换成相反方向的水平恒力乙推这一物体，当恒力乙作用时间与恒力甲作用时间相同时，物体恰好回到原处，此时物体的动能为32J。则在整个过程中，恒力甲、乙对物体做的功分别是多少？4.从倾角为θ的斜面上，水平抛出一个小球，小球的初动能为EK0,如图所示，求小球落到斜面上的动能EK。5.一物体从斜面底端以初动能E滑向斜面，返回到斜面底端的速度大小为V，克服摩擦力做的功为2E，若物块以初动能2E滑向斜面，则（）A.返回斜面底端时的动能为EB.返回斜面底端时的动能为23EC.返回斜面底端时的速度大小是2VD.返回斜面底端时的速度大小为v2F/NCBA1512000400V1/s.m-1OCO·yxXHRABHθxyxXHCθ中小学教育资源站（)，百万资源免费下载，无须注册！中小学教育资源站版权所有6.如图所示，位于竖直平面内的光滑圆轨道，由一段斜的直轨道与之相切的圆形轨道连接而成，圆形轨道的半径为R。一个质量为m的小物块从斜轨道上某处由静止开始下滑，然后沿圆形轨道运动。要求物块能通过圆形轨道的最高点，且在该最高点与轨道间的压力不能超过5mg（g为重力加速度）。求物块初始位置相对于圆形轨道底部的高度h的取值范围。7.一种叫做“蹦极”的现代运动，可以用下面的实验来进行模拟，如图所示，在桌边安一个支架，在支架横臂的端点系上一根橡皮绳，其重力可不计，劲度系数为k，橡皮绳的弹力与其伸长的长度成正比。橡皮绳另一端系一个质量为m的小球，使小球从支架横臂高处由静止下落，小球落到最低点时，便又被橡皮绳拉回然后再落下······已知橡皮绳的弹性势能221KXEP,式中k为劲度系数，x为橡皮绳的伸长量或压缩量。若小球下落的最大高度是L，试求橡皮绳的自然长度？8.一个质量m=0.2kg的小球系与轻质弹簧的一端，且套在光滑竖直的圆环上，弹簧的上端固定于环的最高点A，环的半径R=0.5m,弹簧的原长L0=0.5m,劲度系数为4.8N/m，如图所示，若小球从图中所示的位置B点由静止开始滑动到最低点C时，弹簧的弹性势能EP弹=0.6J。求：小球到C点的速度vc的大小。9.如图所示,倾角为θ的光滑斜面上方有两个质量均为m的小球A、B，两小球用一根长为L的轻杆相连，下面的B求离斜面底端的高度为h，两球从静止开始滑下斜面后进入光滑平面（不计与地面碰撞时的机械能损失）求：（1）两球在光滑平面上运动时的速度（2）在这过程中杆对A求所做的功（3）试分析在哪个过程杆对小球A做功了10.如图所示，水平轨道AB与放置在竖直平面内的1/4圆弧轨道相连，圆弧轨道B端的切线沿水平方向。一个质量m=1.0kg的滑块（可视为质点），在水平恒力F=5.0N的作用下，从A点由静止开始运动，已知A、B之间的距离S=5.5m,滑块与水平轨道间的动摩擦因数μ=0.10，圆弧轨道的半径R=0.30m，取g=10m/s2。（1）求当滑块运动的位移为2.0m时的速度大小;（2）当滑块运动的位移为2.0m时撤去力F，求滑块通过B点时对圆弧轨道的压力大小；（3）滑块运动运动的位移为2.0m时撤去力F后，若滑块恰好能上升到圆弧的最高点，求在圆弧轨道上滑块克服摩擦力所做的功。hmRLABCRo600θBhθhθAhθLBhθFACBOR中小学教育资源站（)，百万资源免费下载，无须注册！中小学教育资源站版权所有答案1.（1）AB段匀加速，BC段做加速度减小的加速C点，车速达到最大，有CFVP0·····························（1）F=f···································（2）由（1）（2）得P0=FvC=400×15w=6×103w(2)对AB段由牛顿第二定律有F-f=ma·································（3）得出a=2m/s2(3)B点的速度VB=P0/F=3m/s···································（4）因此当车速为2m/s时车在做匀加速故由V=at,得t=1s················································（5）2.由动能定理021)2(2cmvmghRhmg······································（1）由平抛得X=Vct·······························································（2）Y=1/2gt2·····························································（3）几何关系得H=X·································································（4）Y+H=2R·····························································（5）由（1）（2）（3）（4）（5）得mH)51(3.设力甲作用时，物体的末速度为V1,力乙作用时，物体的末速度为V2且两段位移大小相等为S，时间相等，因此由平均速度公式（设V1的方向为正方向）力甲作用时：tVS201··········································（1）力乙作用时：tVVS221·······································（2）由（1）（2）得V2=2V1····················································（3）对两过程分别用动能定理有02121mvW甲··································（4）21222121mvmvW乙·······································（5）而2221mv=32J··············································（6）由（3）（4）（5）（6）得W甲=8JW乙=24J中小学教育资源站（)，百万资源免费下载，无须注册！中小学教育资源站版权所有4.由动能定理mgy=EK-EK0······································（1）x=v0t·············································（2）y=1/2gt2···········································（3）tanθ=y/x··········································（4）由（1）（2）（3）（4）得EK=EK0（1+4tan2θ）·····················（5）5.AD6.由机械能守恒得2212mvmgRmgh··········································（1）通过最高点得条件Rvmmg2··················································（2）由（1）（2）得25Rh·······················································（3）最高点压力小于5mg有RvmNmg2且N5mg得Rvmmg26·················································（4）由（1）（4）得Rh5·······················································（5）由（3）（5）得RhR525·····························（6）7.令小球释放位置为零势能面，橡皮经原长为L0，由机械能守恒2210kxmgL············································（1）xLL0···················································（2）由（1）（2）得kmgLLL20·······························（3）8.令C点所在平面为零势能面，B点到C点的竖直距离为h，由机械能守恒弹PcEmvmgh221··········································（1）由几何关系060cosRRh··································（2）由（1）（2）得smvc/3·····································（3）9.对A、B组成得系统机械能守恒KPEE中小学教育资源站（)，百万资源免费下载，无须注册！中小学教育资源站版权所有2)(21)sin(VmmghmlhgmBABA···································（1）mA=mB···································································（2）得sin2glghv·····················································（3）对A由动能定理得221)sin(mvwlhmg···············································（4）