题目

A 0.60-kg basketball is dropped out of a window that is 6.1m above the ground.The ball is caught by person whose hands are 1.5m above the ground.
A)How much work is done on the ball by its weight?What is the gravitational potential energy of the basketball,relative to the ground,when it is
B)Released and
C)Caught?
D)How is the change (Final PE – initial PE) in the ball’s gravitational potential energy related to the work done by its weight?
在work and energy
宝贝鱼上的LJ翻译:
0.60kg篮球退出是6.1m在地面之上的窗口.球由手是1.5m在地面之上的人拿到.
A) 多少工作完成在球由它的重量?什么是篮球的重心势能,相对地面,当它是
B) 发布和
C) 捉住?
D) 变化(最后的PE –最初的PE)在ball’s重心潜在相关能源上对工作由它的重量怎么样?
只能恭参考.

提问时间：2020-10-22

答案

从6.1米高的地方把0.6kg的篮球扔下去v=0
1个人在1.5m的地方接住球
A:重力做工多少?地面是0势点,重力势能是多少?
W=FL=0.6*10*(6.1-1.5)=6*4.6=27.6J
Ep初=6.1*6=36.6J(B)释放点
Ep1.5=1.5*6=9J(C)抓住的地方
结论:Ep初-Ep末=W重力
当外力不做功的时候
D(Ep初-Ep末和W重力有什么关系?)