2021年高三上学期第六次周考物理试题 含答案

2021年高三上学期第六次周考物理试题 含答案

一、选择题(每小题6分，共48分)

1．运动员在110米栏比赛中，主要有起跑加速、途中匀速跨栏和加速冲刺三个阶段，运动员的脚与地面间不会发生相对滑动．以下说法正确的是( )

A ．加速阶段地面对运动员的摩擦力做正功

B ．匀速阶段地面对运动员的摩擦力做负功

C ．由于运动员的脚与地面间不发生相对滑动，所以不论是加速阶段还是匀速阶段，地面对运动员的摩擦力始终不对运动员做功

D ．无论是加速阶段还是匀速阶段，地面对运动员的摩擦力始终做负功

图D5-1

2．xx·河北石家庄质检图D5-1为汽车在水平路面上启动过程中的速度图像，Oa 为过原点的倾斜直线，ab 段表示以额定功率行驶时的加速阶段，bc 段是与ab 段相切的水平直线．下列说法正确的是( )

A ．0～t 1时间内汽车做匀加速运动且牵引力的功率恒定

B ．t 1～t 2时间内汽车牵引力做的功为12mv 22－12mv 2

1

C ．t 1～t 2时间内汽车的平均速度为1

2

(v 1＋v 2)

D ．在全过程中，t 1时刻汽车的牵引力及其功率都是最大值，t 2～t 3时间内牵引力最小

图D5-2

3．xx·辽宁五校协作体模拟如图D5-2所示，A 放在固定的光滑斜面上，B 、C 两小球在竖直方向上通过劲度系数为k 的轻质弹簧相连，C 球放在水平地面上．现用手控制住A ，使细

线刚刚拉直但无拉力作用，并保证滑轮左侧细线竖直、右侧细线与斜面平行．已知A的质量为4m，B、C的质量均为m，重力加速度为g，细线与滑轮之间的摩擦不计，开始时整个系统处于静止状态．释放A后，A沿斜面下滑至速度最大时C恰好离开地面．下列说法正确的是( )

A．斜面的倾角α＝60°

B．A获得的最大速度为2g m 5k

C．C刚离开地面时，B的加速度最大

D．从释放A到C刚离开地面的过程中，A、B两小球组成的系统机械能守恒

图D5-3

4．xx·芜湖模拟一环状物体套在光滑水平直杆上，物体能沿杆自由滑动，绳子一端系在物体上，另一端绕过定滑轮，用大小恒定的力F拉着，使物体沿杆自左向右滑动，如图D5-3所示．物体在杆上通过a、b、c三点时的动能分别为E a、E b、E c，且ab＝bc，滑轮质量和摩擦均不计，则下列关系中正确的是( )

A．E b－E a＝E c－E b B．E b－E a

C．E b－E a>E c－E b D．E a

图D5-4

5．xx·温州八校联考如图D5-4所示，质量为m的滑块以一定初速度滑上倾角为θ的固定斜面，同时施加一沿斜面向上的F＝mg sin θ的恒力；已知滑块与斜面间的动摩擦因数μ＝tan θ，取出发点为参考点，能正确描述滑块运动到最高点过程中产生的热量Q、滑块动能E k、机械能E随时间t及势能E p随位移x变化关系的是图D5-5中的( )

图D5-5

图D5-6

6．xx·厦门5月适应考试如图D5-6所示，两个质量均为m且用轻弹簧相连接的物块A、B放在一倾角为θ的光滑斜面上，系统静止．现在用一平行于斜面向上的恒力F拉物块A，

使之沿斜面向上运动，当物块B 刚要离开固定在斜面上的挡板C 时，物块A 运动的距离为d ，瞬时速度为v ，已知弹簧劲度系数为k ，重力加速度为g ，则( )

A ．此时物块A 运动的距离d ＝mg sin θ

2k B ．此时物块A 的加速度为a ＝

F －kd －mg sin θ

m

C ．此过程中弹簧弹性势能的改变量ΔE p ＝0

D ．经过程中弹簧弹性势能的改变量Δ

E p ＝Fd －12

mv 2

图D5-7

7．xx·苏州一中质检如图D5-7所示，半径为R 、圆心角为60°的光滑圆弧槽固定在高为h 的平台上，小物块从圆弧槽的最高点A 由静止开始滑下，滑出槽口B 时速度方向水平向左，小物块落在地面上C 点，B 、C 两点在以O 2点为圆心的圆弧上，O 2在B 点正下方地面上，则( )

A ．4R ＝h

B ．2R ＝h

C ．R ＝h

D ．R ＝2h

图D5-8

8．xx·潍坊一摸如图D5-8所示，足够长的粗糙斜面固定在水平面上，物块a 通过平行于斜面的轻绳跨过光滑轻滑轮与物块b 相连，b 的质量为m .开始时，a 、b 均静止且a 刚好不受斜面摩擦力作用．现对b 施加竖直向下的恒力F ，使a 、b 做加速运动，则在b 下降h 高度过程中( )

A ．a 的加速度为F m

B ．a 的重力势能增加mgh

C ．绳的拉力对a 做的功等于a 的机械能的增加量

D ．F 对b 做的功与摩擦力对a 做的功之和等于a 、b 的动能的增加量 二、实验题(16分)

9．xx·西城一模利用如图D5-9甲所示的装置做“验证机械能守恒定律”的实验．

图D5-9

(1)除打点计时器(含纸带、复写纸)、交流电源、铁架台、导线及开关外，在下面的器材中，必须使用的还有________(选填器材前的字母)．

A．大小合适的铁质重锤B．体积较大的木质重锤

C．刻度尺D．游标卡尺E．秒表

(2)图乙是实验中得到的一条纸带．在纸带上选取三个连续打出的点A、B、C，测得它们到起始点O的距离分别为h A、h B、h C.重锤质量用m表示，已知当地重力加速度为g，打点计时器打点的周期为T.从打下O点到打下B点的过程中，重锤重力势能的减少量|ΔE p|＝________，动能的增加量ΔE k＝________．

三、计算题(46分)

10．(22分)xx·河南示范性高中五校联考如图D5-10所示，光滑水平面AB与竖直面内的半圆形导轨在B点相切，半圆形导轨的半径为R.一个质量为m的物体将弹簧压缩至A点后由静止释放，在弹力作用下物体获得某一向右的速度后脱离弹簧，当物体经过B点进入导轨的瞬间对轨道的压力为其重力的8倍，之后物体向上运动恰能到达最高点C.不计空气阻力，试求：

(1)物体在A点时弹簧的弹性势能；

(2)物体从B点运动至C点的过程中产生的内能．

图D5-10

11．(24分)xx·山东实验中学二模如图D5-11所示，有一个可视为质点的质量为m＝1 kg 的小物块从光滑平台上的A点以v0＝2 m/s的初速度水平抛出，到达C点时，小物块恰好沿C 点的切线方向进入固定在水平地面上的光滑圆弧轨道，最后小物块滑到紧靠轨道末端D点的质量M＝3 kg的长木板上．已知木板上表面与圆弧轨道末端切线相平，木板下表面与水平地面之间光滑，小物块与长木板间的动摩擦因数μ＝0.3，圆弧轨道的半径为R＝0.4 m，C点和圆弧的圆心连线与竖直方向的夹角θ＝60°，不计空气阻力，g取10 m/s2.

(1)求小物块到达圆弧轨道末端D点时对轨道的压力．

(2)要使小物块不滑出长木板，木板的长度L至少为多大？

图D5-11

参考答案

1．C 因运动员的脚与地面间不发生相对滑动，故地面对运动员的静摩擦力对运动员不做功，选项C正确．

2．D 在0～t1时间内，牵引力恒定，速度均匀增大，由P＝Fv知，牵引力功率也增大，

选项A错误；t1～t2时间内，根据动能定理知W F－W f＝1

2

mv22－

1

2

mv21，选项B错误；由于汽车在

t 1～t 2时间内是做加速度逐渐减小的变加速直线运动，故平均速度v >1

2

(v 1＋v 2)，选项C 错误；

全过程中，t 1时刻牵引力最大，牵引力的功率达到额定功率，之后牵引力功率不变，但牵引力减小，直至牵引力F ＝f ，此后汽车做匀速运动，选项D 正确.

3．B A 、B 两球的速度大小时刻相等，当A 沿斜面下滑至速度最大时，B 球竖直上升的速度也达到最大，此时A 、B 两球的加速度均为零，选项C 错误；根据“此时C 恰好离开地面”可知，弹簧的弹力大小等于mg ，对B 由平衡条件可得，绳的拉力大小为2mg ，对A 由平衡条件得4mg sin α＝2mg ，解得α＝30°，选项A 错误；从释放A 到A 沿斜面下滑至速度最大的过程中，弹簧由被压缩

mg k 逐渐变成被拉伸mg

k

，在此过程中，由三个小球和弹簧组成的系统机械能守恒，设A 获得的最大速度为v ，则4mg ·2mg k ·sin 30°－mg ·2mg k ＝12×5mv 2

，解得v ＝

2g

m

5k

，选项B 正确；从释放A 到C 刚离开地面的过程中，弹簧对B 的弹力先做正功后做负功，所以由A 、B 两小球组成的系统的机械能先增大后减小，即机械能不守恒，选项D 错误．

4．CD 设绳对物体的拉力为T ，力F 与T 大小相等．T 在对物体做功的过程中大小虽然不变，但其方向时刻在改变，因此该问题是变力做功的问题．但是在滑轮的质量以及滑轮与绳间的摩擦不计的情况下，由W ＝Fl cos α及ab ＝bc 知W ab >W bc ，根据动能定理知E b －E a >E c －

E b ，C 正确，A 、B 错误；由a 经过b 到c ，拉力一直做正功，故物体的动能一直在增加，选项

D 正确．

5．C 根据滑块与斜面间的动摩擦因数μ＝tan θ可知，滑动摩擦力等于滑块重力沿斜面向下的分力．施加一沿斜面向上的F ＝mg sin θ的恒力后，滑块的机械能保持不变，重力势能随位移x 均匀增大，选项C 正确，选项D 错误．产生的热量Q ＝fx 随位移x 均匀增大，滑块的动能E k 随位移x 均匀减小，选项A 、B 错误.

6．C 系统静止时，A 受到弹簧平行于斜面向上的弹力kx 1＝mg sin θ，B 即将离开挡板时，B 受到弹簧平行于斜面向上的弹力kx 2＝mg sin θ，物块A 移动的距离d ＝x 1＋x 2＝2mg sin θ

k

，选项A 错误；对A 应用牛顿第二定律得F －kx 2－mg sin θ＝ma ，解得a ＝

F －kx 2－mg sin θ

m

，选项B 错误；由上述讨论知x 1＝x 2，所以弹簧弹性势能的改变量为零，选

项C 正确，选项D 错误.

7．B 小物块从圆弧槽的最高点A 由静止开始滑下，由动能定理有mgR (1－cos 60°)＝12mv 2.滑出槽口B 后做平抛运动，有h ＝12

gt 2

，h ＝vt ，联立解得2R ＝h ，选项B 正确.

8．BD 设斜面倾角为θ，开始时，根据平衡条件得m a g sin θ＝mg ；施加恒力F 时，有

F ＋mg －T ＝ma ，T －m a g sin θ－f ＝m a a ，联立解得a ＝

F －f m a ＋m

m

，选项A 错误；b 下降h 高度过程中，b 的重力势能减少ΔE p 减＝mgh ，a 的重力势能增加ΔE p ＝m a gh sin θ＝mgh ，选项B 正确；

b 下降h 高度过程中，a 受重力、绳的拉力、支持力及滑动摩擦力作用，根据能量守恒定律得，

绳的拉力对a 做的功大于a 的机械能的增加量，选项C 错误；对a 、b 组成的系统应用能量守恒定律得，F 对b 做的功与摩擦力对a 做的功之和等于a 、b 的机械能的增加量，而b 的重力势能的减少量等于a 的重力势能的增加量，所以F 对b 做的功与摩擦力对a 做的功之和等于a 、

b 的动能的增加量，选项B 、D 正确.

9．(1)AC (2) mgh B m （h C －h A ）2

8T

2

(2)打B 点时重锤的速度v B ＝

h C －h A 2T ，从O 到B ，重锤动能的增加量ΔE k ＝12

mv 2

B ＝m （h

C －h A ）2

8T

2

，重锤重力势能的减少量||ΔE p ＝mgh B . 10．(1)7

2

mgR (2)mgR

(1)设物体在B 点时的速度为v B ，受到的弹力为F N B ，则有

F N B －mg ＝m v 2B

R

其中F N B ＝8mg

物体从A 点运动到B 点的过程中，由能量守恒定律可知E p ＝12mv 2

B

解得E p ＝7

2

mgR .

(2)设物体在C 点时的速度为v C ，由牛顿第二定律得

mg ＝m v 2C

R

物体从B 点运动到C 点的过程中，由能量守恒定律得

Q ＝12

mv 2B －? ??

??12

mv 2C ＋mg ·2R

解得Q ＝mgR .

11．(1)60 N ，方向竖直向下 (2)2.5 m (1)小物块在C 点时的速度为

v C ＝

v 0

c os 60°

小物块从C 到D 的过程中，由动能定理得

mgR(1－cos 60°)＝1

2

mv2D－

1

2

mv2C

解得v D＝2 5 m/s

小物块在D点时，由牛顿第二定律得

F N－mg＝m v2D R

解得F N＝60 N

由牛顿第三定律得F′N＝F N＝60 N，方向竖直向下．

(2)设小物块刚滑到木板左端时恰与木板达到共同速度v，小物块在木板上滑行的过程中，小物块与长木板的加速度大小分别为

a1＝μmg

m

＝μg

a2＝μmg M

则有v＝v D－a1t＝a2t

对小物块和木板组成的系统，由能量守恒定律得

μmgL＝1

2

mv2D－

1

2

(m＋M)v2

解得L＝2.5 m.30116 75A4 疤j40470 9E16 鸖28776 7068 灨] x`33157 8185 膅23056 5A10 娐K39492 9A44 驄39577 9A99 骙FW

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