问题 多选题

如图是某中学科技小组制作的利用太阳能驱动小车的装置.当太阳光照射到小车上方的光电板,光电板中产生的电流经电动机带动小车前进.若小车在平直的水泥路上从静止开始加速行驶,经过时间 t 前进距离 s,速度达到最大值 vm,设这一过程中电动机的功率恒为P,小车所受阻力恒为F,那么(  )

A.这段时间内小车先加速运动,然后匀速运动

B.这段时间内电动机所做的功为 Pt

C.这段时间内电动机所做的功为

1
2
m
v2m

D.这段时间内电动机所做的功为FS+

1
2m
v2m

答案

A、小车电动机的功率恒定,速度不断变大,根据功率与速度关系公式P=Fv可知,牵引力不断减小,

根据牛顿第二定律,有

P
v
-F=ma

故小车的运动是加速度不断减小的加速运动,故A错误;

B、这一过程中电动机的功率恒为P,所以W=Pt,

所以这段时间内电动机所做的功为Pt.故B正确.

C、对小车启动过程,根据动能定理,有

W-FS=

1
2
m
v2m

这段时间内电动机所做的功为W=FS+

1
2
m
v2m
,故C错误,D正确.

故选BD.

阅读理解

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