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12-4-5八年级期末考点训练3

12-4-5八年级期末考点训练3
12-4-5八年级期末考点训练3

力学压轴题

重点:力学压轴题考查的重点应该是杠杆、斜面、功、功率的计算,还有浮力、压强的综合计算;

特点:压轴题目一般考理论联系实际的方面较多,题目背景叙述文字多,读懂题意是关键,能从题目中提炼出有用信息;对学生的能力要求是能从生产工具等实物中抽象出杠杆、斜面、连通器等物理模型,运用物理基本规律、基本原理解题;

题型一:简单机械杠杆类

1.如图甲所示,我国“向阳红09”母船正在进行投放“蛟龙号”载人深潜器试验,调控两根伸缩杆,使“∩”形架绕固定在甲板上的转轴转动至图甲所示位置时停止,然后由固定在“∩”形架顶部的绞车通过钢缆释放深潜器.

处在图甲所示位置的“∩”形架可以简化为图乙中以O为支点的杠杆,OB表示“∩”形架,CD表示两根伸缩杆,O到CD的距离为2m,OD、OE的长度分别为3m、6m.不考虑“∩”形架、两根伸缩杆和绞车的重力.

(1)请在图乙中画出投放深潜器时“∩”形架的动力和阻力示意图,此杠杆为杠杆(省力、费力);图乙中,如果保持OB和D点位置不变,调整伸缩杆长度使点C向O点靠近,则两根伸缩杆所承受的拉力(变小、变大、不变)。

(2)深潜器的质量为25t,总体积为24.5m3.在绞车缓慢向下释放深潜器至刚浸没的过程中,两根伸缩杆的最小拉力为多少牛?

2、如图所示,轻质杠杆OP长1m,能绕O点转动,P端用细绳悬于N点。现有一质量为1kg的物体A通过滑环挂在M点(滑环和绳子的质量可忽略),OM的长度为0.1m,由于杆OP与水平方向成30度角倾斜,滑环刚好能由M向P端匀速滑动,摩擦力力为1N,滑动速度为0.02m/s。(g取10N/kg)求:(6分)

⑴若细绳能承受的最大拉力为9N,滑环从M点开始滑动,经过多长时间后细绳会断裂;从滑环自M点滑动到细绳PN断裂时,A所受重力做功的功率;

⑵若细绳不断裂,当滑环滑到P点后,用一沿PO方向的力F拉着滑环匀速返回到M点,拉力F做的功为多少?拉力F为多大?

3.(2012年4月调考杠杆与克服重力做功)平推杠铃架是一种健身器材,结构如图所示,其中直杆ABCD 是质量分布均匀的圆柱体状杠铃杆:训练时人平躺在长凳上,双手将杠铃杆反复向上推起,起到锻炼身体的作用。

(1)请在图中画出杠铃杆的重力的示意图。

(2)杠铃杆两侧可以装配杠铃片,杠铃片按质量可分为5kg、15kg、20kg、25kg共5种型号。往空的杠铃杆

上安装杠铃片时,先将一个杠铃片套在杠铃杆的一端,再将这个杠铃片水平向里推到图示位置,然后在杠铃杆的另一侧安装一个相同的杠铃片。试通过计算说明,在上述安装过程中,为了不使杠铃杆翻转,杠铃杆的质量至少应为多大?

(3)某人在一次健身时,20s内向上推举了10次杠铃杆,已知杠铃杆和杠铃片的总质量为80kg,如果每次

推举时手斜向上移动了50cm,杠铃杆上升的高度为40cm,那么他在这次健身时推举做功的功率约为多少?

4.(2012年元月调考。杠杆与克服重力做功的计算)图甲是电工师傅们挖的埋水泥电线杆的坑,其深度为2.0m,坑口长为2.3m,坑底长为0.3m,坑的宽度仅比电线杆的粗端直径稍大一点。图乙是电工师傅们准备将电线杆埋入坑中时的示意图。

(1)坑中沿长方形的长边方向有一从地面直达坑底的斜坡,这样做的好处是不必将电线杆从底端开始向上竖起,重心上升高度较,施工时既省力又省功。

(2)水泥电线杆的重力为1.0×104N,长为10.0m.其重心O距粗端的距离为4.0m。若按图乙所示的方式竖起电线杆。则刚抬起电线杆的细端时至少需要多大的力?请在图乙中画出动力臂l1和阻力臂l2。

(3)竖起电线杆的过程中,工人师傅们至少要对电线杆做多少功?

题型二:功、功率计算类

5.下图是八浮子(编号为1-8)波浪发电机的俯视图。当浮子随着海面波浪向上浮动时,会通过传动装置带动发电机发电。已知每个浮子的重力为20N,体积为0.016m3,若它在水中上浮带动发电机发电时,露出水面的体积为0.004m3,则:

(1)每个浮子上浮带动发电机发电时,受到的浮力是多少牛?

(2)当海浪的波高为3m、起伏周期为6s时,假设浮子沿竖直方向运动,平均每秒每个浮子带动发电机发电时匀速上浮的高度为0.8m,传动装置和发电机的总效率为30%,求:

①每个浮子对与之相连的硬杆产生的竖直向上的推力是多少牛?

②波浪发电机的输出功率是多少瓦?

题型三:压强浮力综合计算类

6、图甲为加油站加油系统示意图,油罐容积为30m3,为了降低油品损耗,保证油罐运行安全,在油罐顶部安装有调节油罐内外压力平衡的呼吸阀.其中吸气阀(图乙)管口面积为2×10-3m2,重力不计的吸气阀盖在0.6N的恒力作用下堵住吸气管口.加油时,加油机内电动油泵自动启动,加油枪流量控制为Q=5×10-4m3/s;(ρ油取0.8×103kg/m3,外界大气压取105Pa,g=10N/kg)(9分)

(1)某次加油时,加油枪口高出油罐油面3m,加油1分钟耗电900J,则加油系统的效率为多少?(不考虑油面的变化)

(2)若加油枪口横截面积为4×10-4m2,则加油时油在油枪口处流速是多少米/秒?

(3)若油罐中油的体积为10.04m3,罐内气体压强与其体积关系如下表所示(吸气阀未打开前气体质量、温度保持不变),(1)由表可知:当气体质量、温度保持不变时,气体的体积越,压强越;(2)则在加油多长时间后,吸气阀会立即打开?

油罐的气体体积(m3)…19.96 19.98 20 20.02 20.04 …

油罐的气体压强(Pa)…99900 99800 99700 99600 99500 …

7、如图甲所示,放在水平地面上的物体A受到水平向右的力F的作用.力F的大小以及物体A的运动速度大小v随时间t的变化情况如图乙所示.

(1)当t=6s时,物体A受到的摩擦力f的大小为 N,方向为。

当t=7s时,物体A受到的摩擦力f的大小为 N.

(2)根据图乙有关信息,请用公式P=Fv,求F=10N时该力的功率.

(3)如图丙所示,在A的两侧分别挂上柱状重物B,C,且C的一部分浸入水中.已知G B=20N,G C=50N,C的横截面积为30cm2,长度足够,水够深.则当物体A受6N摩擦力作用时,C的下底面受到的水的压强是多少?若物体A移动就会触发报警装置(图中未画出),当物体A不移动时,最高水位与最低水位的差是多少?(g取10N/kg).(6分)

答案:

1、(1)费力,变大。

(2)最小力为15000N

2、(1)40s; 0.1W

(2)拉力做功为W=Gh+fs=0.45m×10N+1N×0.9m=5.4J

F=W/s=5.4J/0.9m=5N

3、(1)略

(2)20kg

(3)160W

4、(1)低

(2)2500N

(3)20000J

5.(1)120N

(2)100N 192W

6.(1)80%

(2)1.25m/s

(3)大、小;80s

7.(1)12,向左,10

(2)0或50W

(3)摩擦力向右时,P=1.2×104 Pa

摩擦力向左时P=8×103Pa

最高水位与最低水位的差为0.8m

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