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人教版必修一《弹力》课时检测卷及答案

人教版必修一《弹力》课时检测卷及答案
人教版必修一《弹力》课时检测卷及答案

物理·必修1(人教版)

第二课时弹力

水平测试

1.(双选)关于形变,下列说法正确的是( )

A.物体形状的改变叫弹性形变

B.物体受到外力作用后发生的形状改变叫弹性形变

C.物体在外力去掉后能恢复原状的形变叫弹性形变

D.任何物体在外力的作用下都能发生形变,不发生形变的物体是不存在的

解析:物体的形变分为弹性形变和非弹性形变,二者区别在于去掉外力后形变是否恢复,能够恢复原状的称为弹性形变,否则为非弹性形变,综上所述选项C、D正确.

答案:CD

2.一个物体静止在水平桌面上时,下列说法正确的是( )

A.物体对桌面的压力就是该物体的重力

B.物体对桌面的压力使该物体发生形变

C.桌面形变产生对物体的支持力

D.以上说法都不正确

解析:物体的重力是对桌面产生压力的原因,但重力不是压力.桌面对物体的弹力使物体发生形变.答案:C

3.(双选)一辆汽车停在水平地面上,下列关于汽车与地面所受弹力的说法正确的是( )

A.地面受到向下的压力,是因为地面发生了形变

B.地面受到向下的压力,是因为汽车发生了形变

C.汽车受到向上的支持力,是因为地面发生了形变

D.汽车受到向上的支持力,是因为汽车发生了形变

解析:弹力是由于发生弹性形变的物体要恢复原状对跟它接触的物体施加的力的作用,恢复原状的能力,也就是该物体对与它接触的物体产生的弹力,所以一个物体受到的弹力是另一个物体发生形变产生的.汽车停在水平地面上,由于重力的作用,汽车与地面相互挤压,发生形变,产生弹力,地面受到向下的压力是由于汽车发生形变要恢复原状产生的,施力物体是汽车;汽车受到向上的支持力是由于地面发生形变要恢复原状产生

的,施力物体是地面.

答案:BC

4.如图所示,

一根弹性杆的一端固定在倾角为30°的斜面上,杆的另一端固定一个重为2 N的小球.小球处于静止状态,则弹性杆对小球的弹力( )

A.大小为2 N,方向平行于斜面向上

B.大小为1 N,方向平行于斜面向上

C.大小为2 N,方向垂直于斜面向上

D.大小为2 N,方向竖直向上

解析:

答案:D

5.如图所示,弹簧测力计和细线的重力不计,一切摩擦不计,重物的重力G=10 N,则弹簧测力计A和B 的读数分别为( )

A.10 N,20 N B.10 N,10 N

C.10 N,0 D.0,0

解析:弹簧测力计的示数是弹簧测力计一端所受拉力的大小.物体都处于静止状态,拉力大小都等于物体所受重力的大小,B正确.

答案:B

6.如图所示的四个图中,M、N两方形木块处于静止状态,它们相互之间一定没有弹力作用的是( )

解析:D选项中N与M间虽接触但不存在挤压力,故D中M、N之间一定不存在弹力.

答案:D

7.体育课上一学生将足球踢向斜台,如图所示.下列关于足球与斜台作用时斜台给足球的弹力方向的说法正确的是( )

A.沿v1的方向

B.沿v2的方向

C.先沿v1的方向后沿v2的方向

D.沿垂直于斜台斜向左上方的方向

解析:足球与斜台的作用是球面与平面的相互作用,足球所受弹力方向垂直于斜台指向足球,即斜向左上方的方向,故D正确.

答案:D

素能提高

8.如图所示,两木块的质量分别为m1和m2,两轻质弹簧的劲度系数分别为k1和k2,上面木块压在上面弹簧上(但不拴接),整个系统处于平衡状态.现缓慢向上提上面的木块,直到它刚离开上面的弹簧.在这个过程中,下面木块移动的距离为( )

A.m 1g k 1

B.m 2g k 2

C.m 1g k 2

D.m 2g k 1

解析:在没有施加外力向上提时,弹簧k 2被压缩的长度Δx =

1

+m 2

k 2

,在用力缓慢向上提至m 1刚离开

上面弹簧时,弹簧k 2仍被压缩,但压缩量Δx′=m 2g k 2.所以,在此过程中,下面木块移动的距离Δx -Δx′=m 1g

k 2.

故答案为C.

答案:C

9.一轻弹簧原长为10 cm ,在它的下端挂一个重为4 N 的物体时,弹簧的长度为12 cm.若在它的下端挂一个重为8 N 的物体时,弹簧的长度为多少?

解析:弹簧的原长l 0=10 cm ,当弹簧下端的重物静止时,弹簧的弹力等于重物的重力.当挂4 N 的重物时,弹簧的弹力F 1=4 N ,弹簧的长度l 1=12 cm ,

则形变量x 1=l 1-l 0=2 cm ; 当挂8 N 的重物时,F 2=8 N , 设弹簧的长度为l 2,则x 2=l 2-l 0. 由胡克定律F =kx 可知F 1x 1=F 2

x 2,

代入数据解得l 2=14 cm. 答案:14 cm

10.如图所示,一根弹簧没有悬挂重物时,其自由端B 正对齐刻度尺的零度线,挂上100 N 重物时,正对着刻度20,试问:

(1)当弹簧分别挂50 N和150 N重物时,自由端所对刻度尺的刻度是多少?

(2)若自由端所对刻度是18,这时弹簧下端挂了多重的重物?

解析:由实验装置可知,弹簧与刻度尺已经组成了一支弹簧测力计.

(1)当F1=50 N时,正对刻度10;

当F2=150 N时,正对刻度30.

(2)设所挂重物重为G,则G

100 N =

18

20

解得G=90 N.

答案:(1)10 30 (2)90 N

圆柱体的扭转

圆柱体两端受到一对大小相等、方向相反的力偶矩时,将发生扭转形变.在微小扭转形变下,圆柱体的各横截面间距不变,即圆柱体不伸长或缩短;各横截面上的半径仍保持为直线

但发生相对转动.圆柱体两端面相对转过的角度叫圆柱体的扭转角,用Φ表示.如右图所示.从图中可以看出,柱面上每一根母线随着圆柱体的扭转都倾斜一个角度,使得柱面上每一个“正方形”面积元都变成了“菱形”.所以扭转形变本质上是剪切形变,母线转过的角度Ψ就是相应的剪切应变.

在微小形变的条件下,剪切应变Ψ=r Φ

l .r 表示体积元所在半径,l 表示柱长.可见,在同一同心圆薄层内

剪切应变相同,不同层内剪切应变不同,中心轴线处的狭长体积元无剪切应变,圆柱表面上体积元的剪切应变最大.剪切应变Ψ和体积元所在半径r 一定时,杆越长,即l 越大,则扭转角Φ越大.

因内外层剪切应变不同,根据剪切形变的胡克定律,内外层剪切应力也不同,靠外层剪切应力较大.剪切应力的出现起着抵抗扭转形变的作用,因此抵抗形变的任务主要是由外层材料来承担,靠近中心轴线的材料几乎不起什么作用.

经过计算可证明,产生扭转的力偶矩M 和实心圆柱扭转角Φ有如下关系: M =πNR 4

2l

φ=c φ.

R 和l 分别表示圆柱的半径与长度,N 为剪切模量,式中c =πNR

4

2l

称为圆柱体的扭转系数.

当M 一定时,R 越大,l 越小,则Φ越小,即短而粗的圆柱体具有较强的抵抗扭转形变的能力.反之,细而长的圆柱体抵抗扭转形变的能力较弱.

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