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2014高考复习(模拟题汇编):专题六_机械能(4)

2014高考复习(模拟题汇编):专题六_机械能(4)
2014高考复习(模拟题汇编):专题六_机械能(4)

专题六:机械能

1.(2013山东济南外国语学校测试)运动员从悬停在空中的直升机上跳伞,伞打开前可看做是自由落体运动,打开伞后减速下降,最后匀速下落。如果用h表示下落高度,t表示下落的时间,F表示人受到的合力,E表示人的机械能,E p表示人的重力势能,v表示人下落的速度,在整个过程中,如果打开伞后空气阻力与速度平方成正比,则下列图象可能符合事实的是()

【答案】AC

【解析】重力势能逐渐减小,E p=mgH=mg(H0-h),即重力势能与高度是线性关系,故A正确。机械能的变化等于除重力外其余力做的功,故自由落体运动过程机械能守恒,故B错误.运动员在伞打开前可看作是自由落体运动,即空气阻力忽略不计,故只受重力;开伞后减速下降,空气阻力大于重力,故合力向上,当阻力减小到等于重力时,合力为零,做匀速直线运动;即合力先等于重力,然后突然反向变大,且逐渐减小到零,故C正确。由于空气阻力与速度平方成正比,故伞打开后的v-t图像不应该是直线,D错误。

2. (2013山东济南测试)如图所示,质量为m的滑块从斜面底端以平行于斜面的初速度v0冲上固定斜面,沿斜面上升的最大高度为H。已知斜面倾角为α,斜面与滑块间的摩擦因数为μ,且μ

E、

E k、

势能E P与上升高度h之间关系的图象是 ( )

【答案】D

【解析】滑块机械能的变化量等于除重力外其余力做的功,故滑块机械能的减小量等于克服

阻力做的功,故上升阶段 E=E 0-F 阻sin h α,下降阶段 E=E 0′-F 阻sin h

α

,故B 错误;动能的变化量等于外力的总功,故上升阶段-mgh-F 阻sin h

α

=E k -E 0 ,下降阶段 mgh-F 阻

sin h

α

=E k -E 0′,C 错D 对;上升阶段势能E P =mgh, 下降阶段势能E P =mgh ,图象应该都是A 图所示的“上行”段,A 错误。

3.(2013山东师大附中质检)如图所示,一个小孩从粗糙的滑梯上加速滑下,对于其机械能的变化情况,下列判断正确的是

A .重力势能减小,动能不变,机械能减小

B .重力势能减小,动能增加,机械能减小

C .重力势能减小,动能增加,机械能增加

D .重力势能减小,动能增加,机械能不变 【答案】B

【解析】小孩下滑过程中,受到重力、支持力和滑动摩擦力,其中重力做正功,支持力不做功,滑动摩擦力做负功;重力做功是重力势能的变化量度,故重力势能减小;小孩加速下滑,故动能增加;除重力外其余力做的功是机械能变化的量度,摩擦力做负功,故机械能减小;故选B .

4.(2013山东济南期中检测)质量为m 的物体,由静止开始下落,由于阻力的作用,下落的加速度为4g/5,在物体下落高度为h 的过程中,下列说法正确的是( ) A .物体的动能增加了4mgh/5 B .物体的机械能减少了4mgh/5 C .物体克服阻力做功mgh/5 D .物体的重力势能减少了mgh 【答案】ACD

【解析】因物体的加速度为g ,故说明物体受阻力作用,由牛顿第二定律可知,mg-f=ma ;解得f=

15mg ;重力做功W G =mgh ; 阻力做功W f =-1

5mgh ;由动能定理可得动能的改变量△E k =W G +W f =

45mgh ;故A 正确;阻力做功消耗机械能,故机械能的减小量为1

5

mgh ;故B 错误;阻力做功为W f ,则物体克服阻力所做的功为

1

5

mgh ;故C 正确;重力做功等于重力势能的改

变量,重力做正功,故重力势能减小mgh ,故D 正确;故选ACD .

5.(2013上海徐汇区期末)18.如图所示,圆心在O 点、半径为R 的光滑圆弧轨道ABC 竖直固定在水平桌面上,OC 与OA 的夹角为60°,轨道最低点A 与桌面相切. 一足够长的轻绳两端分别系着质量为m 1和m 2的两小球(均可视为质点),挂在圆弧轨道光滑边缘C 的两边,开始时m 1位于C 点,然后从静止释放。则( )

(A )在m 1由C 点下滑到A 点的过程中两球速度大小始终相等

(B )在m 1由C 点下滑到A 点的过程中重力对m 1做功的功率先增大后减少

(C )若m 1恰好能沿圆弧下滑到A 点,则m 1=2m 2 (D )若m 1恰好能沿圆弧下滑到A 点,则m 1=3m 2 答案:BC

解析:m 1位于C 点从静止释放后,m 1沿圆弧轨道下滑,在m 1由C 点下滑到A 点的过程中,

m 1和m 2的两小球和地球组成系统的机械能守恒,m 1速度沿轻绳方向的分速度和m 2速度相等,

选项A 错误;在m 1由C 点下滑到A 点的过程中重力对m 1做功的功率先增大后减少,选项B 正确;若m 1恰好能沿圆弧下滑到A 点,根据机械能守恒定律,m 1gR(1-cos60°)=m 2gR ,解得

m 1=2m 2,选项C 正确D 错误。

6.(2013山东济南期中检测)如图所示,小球a 、b 的质量分别是2m 和m ,a 从倾角为30°的光滑固定斜面的顶端无初速下滑,b 从斜面等高处以初速度v 0平抛,比较a 、b 落地的运动过程有( )

A . a ,b 两球同时到达地面

B . a ,b 落地前的速度相同

C . 重力对a 、b 做的功相同

D . 落地前a , b 两球重力做功的瞬时功率相等 【答案】A

【解析】物体a 受重力和支持力,F 合=2mgsin30°,根据牛顿第二定律,a=g/2.物体b 做平抛运动,加速度为g .知两物体的加速度不变,所以两物体都做匀变速运动,设斜面高度

2

为h ,则21222a g h t =?,2

12

b

h gt =?,解得a b t t =,故A 正确.对a 运用动能定理,2mgh=

122mv a 2

-0,对b 运用动能定理,有mgh=12mv b 2-12

mv 02,知b 球的速率大于a 球的速率.故

B 、

C 错误.落地时a 、b 在竖直方向的速度大小分别为1

2

2g g ,由功率的计算公式P=Fv 可知落地前a 、b 两球重力做功的瞬时功率不相等,D 错误(由选项AC 可知ab 两物体重力做的功不相等,运动时间相等可得出重力的平均功率不相等)。

7.(2013河北正定中学测试)如图所示,质量为m 的物体沿

动摩擦因素为μ的水平面以初速度v 0 从A 点出发到B 点时速度变为v ,设同一物体以初速度v 0从A′点先经斜面A′C,后经斜面CB′ 到B′点时速度变为v′ ,两斜面在水平面上投影长度之和等于AB 的长度,则有( ) A .v′>v B .v′=v C .v′

解析:设AC 在水平面投影长度为L 1,物块沿斜面AC 运动所受摩擦力为f=μmgcos θ,摩擦力做功W1=f L 1/cos θ1=μmg L 1,同理,物块沿斜面BC 运动所受摩擦力为f=μmgcos θ2,摩擦力做功W 2=f L 2/cos θ2=μmg L 2,根据动能定理可知,两种过程到达B 点和B’点,速度相等,选项B 正确。

8.(2013山东济南测试)如图所示,置于足够长斜面上的盒子内放有光滑球B ,B 恰与盒子前、后壁接触,斜面光滑且固定于水平地面上.一轻质弹簧的一端与固定在斜面上的木板P 拴接,另一端与A 相连.今用外力推A 使弹簧处于压缩状态,然后由静止释放,则从释放盒子直至其获得最大速度的过程中 ( )

A .弹簧的弹性势能一直减小直至为零

B .A 对B 做的功等于B 机械能的增加量

C .弹簧弹性势能的减小量等于A 和B 机械能的增加量

D .A 所受重力和弹簧弹力做功的代数和小于A 动能的增加量

【答案】BC

【解析】由于弹簧一直处于压缩状态,故弹性势能一直减小,但没有减为零,故A 错误;除

重力外其余力做的功等于物体机械能的增加量,故A 对B 做的功等于B 机械能的增加量,故B 正确;由于AB 整体和弹簧系统机械能守恒,故弹簧弹性势能的减小量等于A 和B 机械能的增加量,故C 正确;对物体A ,重力、支持力、弹簧弹力和B 对A 弹力的合力做的功等于动能的增加量,故重力和弹簧弹力做功的代数和不等于A 动能的增加量,故D 错误.

9.(2013山东济南期中检测)如图所示,一人站在商场的自动扶梯的水平踏板上,随扶梯一起斜向上做加速运动,则下列说法中正确的是( ) A .人只受到重力和踏板对人的支持力两个力作用 B .人对踏板的压力大于人所受到的重力 C .踏板对人做的功等于人的机械能增加量 D .人所受合力做的功等于人的机械能的增加量 【答案】BC

【解析】人的加速度斜向上,将加速度分解到水平方向和竖直方向得:ax=acos θ,方向水平向右;ay=asin θ,方向竖直向上,水平方向受静摩擦力作用,f=ma=macos θ,水平向右,故A 错误;竖直方向受重力和支持力,FN-mg=masin θ,所以FN >mg ,故B 正确.由机械能守恒定律可知,除重力外的力做的功等于物体机械能的增加,C 正确;由动能定理可知人所受合力做的功等于人的动能的增加量,D 错误。

10.(2013山东师大附中质检)质量为m 的汽车,启动后沿平直路面行驶,如果发动机的功率恒为P ,且行驶过程中受到的摩擦阻力大小一定,汽车速度能够达到的最大值为v ,那么当汽车的车速为v /3时,汽车的瞬时加速度的大小为

A .mv p

B .

mv p 2 C .mv

p 3

D .mv

p 4

【答案】B

【解析】当汽车匀速行驶时,有f=F=

P

v

.根据P=F′3v ,得F′= 3P v ,由牛顿第二定律

得a= 'F f m -= 32P P

P v v m mv

-

=.故B 正确,A 、C 、D 错误。 11.(2013山东师大附中质检)如图所示,用一与水平方向成α的力F 拉一质量为m 的物体,使它沿水平方向匀速移动距离s ,若物体和地面间的动摩擦因数为μ,则此力F 对物体做的功,下列表达式中正确的有

A .Fscos α

B .μmgs

C .μmgs /(cos α-μsin α)

D .μmgscos α/(cos α+μsin α) 【答案】AD

【解析】由功的定义式可得,F 的功为 W=Fscos α,A 正确;对物体受力分析知,竖直方向受力平衡 mg=Fsin α+F N ,摩擦力的大小 f=μF N =μ(mg-Fsin α),由于物体匀速运动,由动能定理得,Fscos α-fs=0,联立以上各式解得F=μmg /(cos α+μsin α),将结果代入W=Fscos α可知选项D 正确。

12(10分). (2013年浙江五校联考)如图甲所示,竖直平面内的光滑轨道由倾斜直轨道AB 和圆轨道BCD 组成,AB 和BCD 相切于B 点,CD 连线是圆轨道竖直方向的直径(C 、D 为圆轨道的最低点和最高点),已知∠BOC =30?。可视为质点的小滑块从轨道AB 上高H 处的某点由静止滑下,用力传感器测出滑块经过圆轨道最高点D 时对轨道的压力为F ,并得到如图乙所示的压力F 与高度H 的关系图象,取g =10m/s 2

。求: (1)滑块的质量和圆轨道的半径;

(2)是否存在某个H 值,使得滑块经过最高点D 后能直接落到直轨道AB 上与圆心等高的点。若存在,请求出H 值;若不存在,请说明理由。 .解析:

(1)mg (H -2R )= 12

mv D 2 1分

F +mg = mv D 2

R

1分

第甲

C

得:F =2mg (H -2R )

R

-mg

取点(0.50m ,0)和(1.00m ,5.0N )代入上式得:

m =0.1kg ,R =0.2m 2分

(2)假设滑块经过最高点D 后能直接落到直轨道AB

OE =R

sin30?

x = OE =v DP t 1分

R =12

gt 2

得到:v DP =2m/s 1分 而滑块过D 点的临界速度

v DL =gR =2 m/s 1分

由于:v DP > v DL 所以存在一个H 值,使得滑块经过最高点D 后能直接落到直轨道AB 上与圆心等高的点 1分

mg (H -2R )= 12

mv DP 2 1分

得到:H =0.6m 1分

13.(15分) (2013湖州联考)如图所示,水平桌面上有一轻弹簧,左端固定在A 点,弹簧

处于自然状态时其右端位于B 点。水平桌面右侧有一竖直放置的光滑轨道MNP ,其形状为半径R =0.8 m 的圆环剪去了左上角135°的圆弧,MN 为其竖直直径,P 点到桌面的竖直距离也是R 。用质量m 1=0.4kg 的物块将弹簧缓慢压缩到C 点,释放后弹簧恢复原长时物块恰停止在B 点。用同种材料、质量为

m 2=0.2kg 的物块将弹簧也缓慢压缩到C 点释

放,物块过B 点后其位移与时间的关系为x=6t-2t 2

,物块从桌面右边缘D 点飞离桌面后,由P 点沿圆轨道切线落入圆轨道。g =10 m/s 2

,求:

(1)BP 间的水平距离;

(2)判断m 2能否沿圆轨道到达M 点;

BA

C

(3)释放后m 2在水平桌面上运动过程中克服摩擦力做的功。

.(15分)解:(1)设物块由D 点以初速度v D 做平抛,落到P 点时其竖直速度为v y ,有 2

y v =2gR,

而 v y = v D tan45°

解得 v D =4m/s 。 (2分)

设平抛运动时间为t ,水平位移为x 1,有2

2

1gt R = t v x D =1 解得x 1=1.6m 。(2分)

由题意可知小球过B 点后做初速度为v 0 =6m/s ,加速度大小为a =4m/s 2

的匀减速运动,减速到v D 。 BD 间位移为x 2 ,有 22202ax v v D =-

所以BP 水平间距为 21x x x +== 4.1m (2分) (2)若物块能沿轨道到达M 点,其速度为M v ,有

12m 2v M 2=1

2

m 2v D 2-2m 2gR (2分) 解得: 2816-=M v < gR

即 物块不能到达M 点 (2分)

(3)设弹簧长为AC 时的弹性势能为E P ,物块与桌面间的动摩擦因数为μ

μm 2g=m 2a , 释放m 1时, E P =μm 1gx CB 释放m 2时, E P =μm 2gx CB +

1

2

m 2v 02,

解得 E P =7.2J 。 (3分)

设m 2在桌面上运动过程中克服摩擦力做功为W f ,有:

E P -W f =

1

2

m 2v D 2,解得:W f =5.6J 。

2014年全国高考英语试题分类汇编:之代词+名词 Word版含解析

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