some physics question

[S]【YU】

epe = 1/2 kx^2
max extension x, constant k, can store max epe = 38920J
as mentioned, 40000J > 38920J, required x exceed elastic limit and deform => cannot restore to original length.
To store more epe per extension, only way is to increase k <= force constant.

aildux

epe = 1/2 kx^2
max extension x, constant k, can store max epe = 38920J
as mentioned, 40000J > 38920J, required x exceed elastic limit and deform => cannot restore to original length.
To store more epe ...
[S]【YU】 發表於 16-4-2009 13:15

既系點?會入水?
假設高度系1000 m
500000 J > 38920 J所以人系1000 m高跳落來依然會入水?(string唔斷的話)

[S]【YU】

好難講,因為過左elastic limit之後,就唔會proportional 咁 extend.

大咁多,應該會過埋yield point...斷得啦...雙腳...

Soldier

唔理條繩係幾時拉長都好,佢既目的係抵消返個人loss PE所gain既KE
去到0m果陣時,個人既KE就係40000J
為左令個人係0m之前停低,條繩就要有大過40000J既elastic PE
e+條繩得38920J既elastic PE,點可以係個人插水之前拉停 ...
卍無極王者卍 發表於 13-4-2009 13:10

我對題目既理解係:當一個人玩笨豬jump既時侯,條繩由just taut到fully extenstion,會store左38920J既energy(gPE-->elastic PE),之後,佢會回彈(條繩唔會斷,38920J只係所表條繩fully extenstion個時所儲既能量),所以,計返有幾多gPE轉到條繩fully extenstion個時既elastic PE(38920J),就知個人有冇掉落水

我堅持自己的答案
aildux 發表於 16-4-2009 10:24

十卜你既答案

其實你係咪打漏d野?
elastic potential energy唔係應該=loss in pe+ loss in ke咩
wilson1993912 發表於 12-4-2009 18:41

佢可能當gPE由橋上開始計算,咁就會包埋KE

kkuukkoo

有冇ans???
冇做黎都冇用