A student who just got 10/10 on a Physics midterm decides to try bungee jumping as a reward.He attaches an elastic bungee cord to his ankles and happily jumps off a tall bridge across a river.He ends up barely touching the water before the cord jerks

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A student who just got 10/10 on a Physics midterm decides to try bungee jumping as a reward.He attaches an elastic bungee cord to his ankles and happily jumps off a tall bridge across a river.He ends up barely touching the water before the cord jerks

A student who just got 10/10 on a Physics midterm decides to try bungee jumping as a reward.He attaches an elastic bungee cord to his ankles and happily jumps off a tall bridge across a river.He ends up barely touching the water before the cord jerks
A student who just got 10/10 on a Physics midterm decides to try bungee jumping as a reward.He attaches an elastic bungee cord to his ankles and happily jumps off a tall bridge across a river.He ends up barely touching the water before the cord jerks him back up.The bungee cord has an unstretched length of l=7.0 m and a spring constant of k=24 N/m; the distance from the bridge to the water surface is L=57 m.Eventually you will be asked to find the mass,m,of the student.
Air resistance can be neglected.
The bungee cord applies a constant (upward) force on the student.
With the assumptions made,the sum of the gravitational potential energy of the student,the kinetic energy of the student,and the potential energy stored in the bungee cord is constant as the student moves.
When the student just steps off the bridge,his kinetic energy is zero.
Even though L is much larger than the height of the student,the student cannot be treated as a particle.
The best way to solve this problem is to use conservation of mechanical energy.

A student who just got 10/10 on a Physics midterm decides to try bungee jumping as a reward.He attaches an elastic bungee cord to his ankles and happily jumps off a tall bridge across a river.He ends up barely touching the water before the cord jerks
通过机械能守恒来算.初始和最终的动能都为0.取水面为零势能面,则初始重力势能为mgL,最终重力势能为0.初始弹性势能为0,最终弹性势能为k(L-l)^2/2.
故有mgL=k(L-l)^2/2
质量m就能算出了.

我英语不好 物理好 不好意思这个题都算错了 不是求质量 是下列哪个选项正确 1 空气阻力可以忽略不计。 2蹦极绳有持续向上的力作用于学生 3随着不断地假设 重力的势能的总和是动能 和储存在蹦极线的势能 4学生刚刚跳落的时候 动能为零 5就算L足够长 学生也不能看做一个质点 6解决这个题目的最好方法是机械能守恒 题目下面有解释 反正就是蹦极之类的 上面哪个是对的啊...

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我英语不好 物理好 不好意思

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说有个学生要蹦极,从桥上跳下来,桥高L,绳子长l,劲度系数24,质量忽略,绑人脚上,人的质量m高度不能忽略,我设他是h…他跳下来,刚好碰到水面,速度为0,动能为0。整个过程机械能守恒,求m…
刚跳下来l高度里,绳子的拉力不做功,之后才开始,一直到人碰到水面结束,设重力加速度g
mgL=mgl+1/2*k*(L-l-h)^2
m=(k(L-l-h)^2)/(2g(L-l))<...

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说有个学生要蹦极,从桥上跳下来,桥高L,绳子长l,劲度系数24,质量忽略,绑人脚上,人的质量m高度不能忽略,我设他是h…他跳下来,刚好碰到水面,速度为0,动能为0。整个过程机械能守恒,求m…
刚跳下来l高度里,绳子的拉力不做功,之后才开始,一直到人碰到水面结束,设重力加速度g
mgL=mgl+1/2*k*(L-l-h)^2
m=(k(L-l-h)^2)/(2g(L-l))
自己把要的数据代进去算

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讲的是一个小孩蹦极的题目,问的是小孩的质量极限,mass应为MAX
下面半段是做法提示,里面说了理想条件和注意要点
要求用机械能转换做
考虑到本题目要求使用机械能转换,有题设可知,假设小孩体重最大,在L=0处,小孩具有最大重力势能,在L=57处,具有最大弹性势能
在L=57处,弹性势能Ek=0.5*k*(L-7)^2=0.5*24*50*50=30000
起...

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讲的是一个小孩蹦极的题目,问的是小孩的质量极限,mass应为MAX
下面半段是做法提示,里面说了理想条件和注意要点
要求用机械能转换做
考虑到本题目要求使用机械能转换,有题设可知,假设小孩体重最大,在L=0处,小孩具有最大重力势能,在L=57处,具有最大弹性势能
在L=57处,弹性势能Ek=0.5*k*(L-7)^2=0.5*24*50*50=30000
起点处重力势能Ep=mgL=Ek=30000
得到m=Ek/gL=30000/(10*57)=44.776kg
即,小孩不应超过44.77公斤
另,题目中涉及到儿童不应考虑为质点,保留其可能为开放式问题的可能性,自行设定小孩身高为x,并将上式中的L全部替换为(L-x)
不包翻译,分数拿来

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