Question

在利用自由落体“验证机械能守恒定律”的实验中， （1）下列器材中不必要的一项是 （只需填字母代号）． A

（2）关于本实验的误差，说法不正确的一项是

A．选择质量较小的重物，有利于减小误差

B．选择点击清晰且第1、2两点间距约为2mm的纸带，有利于减小误差

C．先松开纸带后接通电源会造成较大的误差

D．实验产生误差的主要原因是重物在下落过程中不可避免地受到阻力的作用

(3)如在此实验中，打点计时器所用电源频率为50 Hz，当地重力加速度值为9.80m/s2，测得所用重物的质量为1.00kg。若按实验要求正确地选出纸带进行测量，测量得A、B、C三个计数点到第一个点O的距离如图所示，单位均是cm，那么(计算结果均保留三位有效数字)：

①打点计时器打下计数点B时，物体的速度VB = ____ ____m/s

②从起点O到打下计数点B的过程中重力势能减少量是ΔEP=_________J，此过程中物体动能的增加量是ΔEK=__ __J；

③通过计算，数值上ΔEP ΔEK (填“>”“=”或“<”) ，

原因是

④实验的结论是_______________________________________
点O到A的距离是7.01，O到B的距离12.05，O到C的距离是19.01，需要B点速度的答案
此题是否有误？

Answer 1

解：
（1）
A
（2）
电源频率为50Hz说明每0.02秒打一点，同时落体法验证机械能守恒每个点都是一个计数点
Vb=(OC-OA)/(0.04)=3m/s
ΔEp=mgOB=0.687J
ΔEk=mVb^2/2=4.5J
如果没有计算错，那么应该就是有问题了
假设不受阻力，如果速度到达3m/s
那么2gH=v^2-0，H=0.45m=45cm，就算有误差也不会差那么大吧
要不然你把图发来，可能它这里的计数点选择不一样，那样的话那里的时间就不是0.04s

Answer 2

第二个题的答案：AC

追问

第二题答案是A

追答

看你怎么去验证，要是用B的那种办法，C一定会造出很大误差，要是在纸带上选择两个点来验证，初速度不需要为0就没什么大问题

Answer 3

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