some physics question
Bungy jumping os an exciting sport that started in New Zealand. The person has a strong piece of elastic tied to his ankles.He then jumps from a bridge.Length of the elastic when not stretched=50m
Height of the bridge over water=80m
Mass of the boy=50kg
(a) Find the velocity of the boy when the elastic is just taut.
(b) When the elastic extends,the elastic stores elastic potential energy.When the elastic fully stretched,the loss in potential energy of the bungy jumper is equal to elastic potential energy.At full extension,the elastic potential energy is 38920J. Find the maximum extension of the elastic .Does the boy go into the water?
(c)Different people try bungy jumping.What adjustment should be made to the apparatus? 其實你係咪打漏d野?
elastic potential energy唔係應該=loss in pe+ loss in ke咩 本帖最後由 aildux 於 16-4-2009 10:16 編輯
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 http://www.nakuz.com/bbs/images/common/back.gif
elastic potential energy=loss in PE(mgh'=50*10*e)+ loss in KE
loss in KE(mv^2/2)=loss in PE(mgh=50*10*50)
a)v^2/2=gh
v^2=2*10*50
v=31.6 ms^-1
b)38920=mgh'
h'=38920/(50)(10)
=77.84 m
<80 m
.'.the boy won't go into the water
the maximum extension is 77.84-50=27.84 m
c)use a string with higher force constant instead 本帖最後由 卍無極王者卍 於 12-4-2009 22:46 編輯
elastic potential energy=loss in PE(mgh'=50*10*e)+ loss in KE
loss in KE(mv^2/2)=loss in PE(mgh=50*10*50)
a)v^2/2=gh
v^2=2*10*50
v=31.6 ms^-1
b)38920=mgh'
h'=38920/(50)(10)
...
aildux 發表於 12-4-2009 19:53 http://www1.nakuz.com/bbs/images/common/back.gif
b part錯吧
個人係80m既gravitational PE有40000J
條繩得果38920J既elastic PE
佢就算拉到盡都抵消唔到個人既PE啦......
所以個人應該係會插左入水吧 條繩係50m先可以拉長bor~
唔係應該38920=10*50*30-10*50*h
咩
但我計到個h係-ve-.- 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佢呢? 本帖最後由 7ond 於 14-4-2009 18:20 編輯
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 http://www.nakuz.com/bbs/images/common/back.gif
哦我明喇 -.-"
我計到full extension係27.84 ge -.-?
到底個elastic potential energy = loss in pe 定 = loss in pe + loss in pe -.-?
其實有冇答案.. 本帖最後由 LSM 於 14-4-2009 14:36 編輯
好似係
PE(50m)+PE(h')=38920
之後h'如果大過30 就入水 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 http://nakuz.com/bbs/images/common/back.gif
但佢跌落黎果陣會lose左PE架wo
應該要計埋pe 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 http://www.nakuz.com/bbs/images/common/back.gif
我堅持自己的答案
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