问题 多选题

如图所示,质量m=1kg的物块,以速度v0=4m/s滑上正沿逆时针转动的水平传送带,传送带两滑轮A、B间的距离L=6m,已知传送带的速度v=2m/s,物块与传送带间的动摩擦因数μ=0.2. 关于物块在传送带上的运动,以下说法正确的是(  )

A.物块滑离传送带时的速率为2m/s

B.物块在传送带上运动的时间为4s

C.皮带对物块的摩擦力对物块做功为6J

D.整个运动过程中由于摩擦产生的热量为18J

答案

A、B、滑块先向右匀减速,对于向右滑行过程,根据牛顿第二定律,有

μmg=ma

解得

a=μg=2m/s2

根据运动学公式,有

0=v0-at1

x=

v0+0
2
t1

解得

t1=2s

x=4m

向左匀加速过程,根据运动学公式,有

x1=

v
-0
2a
=
4
2×2
=1m

t2=

v
a
=1s

最后3m做匀速直线运动,有

t3=

x-x1
v
=
4-1
2
=1.5s

即滑块在传送带上运动的总时间为:t=t1+t2+t3=4.5s

故A正确,B错误;

C、向右减速过程和向左加速过程中,摩擦力为恒力,故摩擦力做功为:Wf=-f(x-x1)=-μmg(x-x1)=-6J,即物体克服摩擦力做功6J,或摩擦力对物体做功-6J,故C错误;

D、整个运动过程中由于摩擦产生的热量等于滑块与传送带之间的一对摩擦力做功的代数和,等于摩擦力与相对路程的乘积;

物体向右减速过程,传送带向左移动的距离为:l1=vt1=4m

物体向左加速过程,传送带运动距离为:l2=vt2=2m

即Q=f•S=μmg•[(l1+x)+(l2-x1)]=18J,故D正确;

故选AD.

单项选择题
阅读理解

阅读理解。

     We are all called upon to make a speech at some point in life, but most of us don't do a very good job.

This article gives some suggestions on how to give an effective speech.

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