Work done

2007-12-20 2:46 am
一樣10N object 如果向下move 1m ,向上move 1m
雖displacement 是0
咁係咪應該咁諗
work done when moving up= mgh = 10J
work done when moving down = mgh =10J
Total work done = 2 mgh =20J
咁係咪即係FS 個 S 不等於net displacement
如果好似呢個case 就要分開兩個motion 去計
THX~~

回答 (3)

2007-12-20 3:03 am
✔ 最佳答案
唔係 如果你take upwards係positive
咁 work done when moving up= mgh = 10*10*10= 1000J /1kj
work done when moving down = mgh = -10*10*10J = -1000J / -1kj
Total work done = 兩個相加 = 0
FS 係個s係displacement
同埋你意思係咪10kg object mass ge單位係kg

2007-12-19 19:09:10 補充:
其實唔洗分兩個case去計 你可以直接先揾displacement 再用work done=Fs計s = 10 (-10) = 0so by W=Fs W=0J

2007-12-19 19:12:12 補充:
= = 樓下果位,, CE answer都係咁答

2007-12-19 19:15:13 補充:
according to ce 課程work done = F*distance 係out of syllabus
2007-12-21 3:27 am
WD= Fs 係in ce syllabus架
我好肯定
你向上就gain PE=10 x 1=10J
你向下就lost PE=-(10 x 1)=-10J
WD=10+(-10)=0J
你向上1m向下1m,咁番左去starting point
s係displacement而唔係distance
2007-12-20 3:04 am
係呀,呢個case要分開兩個motion 去計

功係由積分所定出黎.
因為積分係單向既,由1點去到另一點.

W=Fs係中學既計法
因為由up轉down,個F既方向係相反左
自然條式就唔岩用,因為用呢條式果陣,係當F係不隨s而變..
用英文既數學黎講就係l:F is not a function of s.

F係你個case既中間轉變過,所以唔可以當呢件事係一個motion

因為W=Fs係當一個過程,所以一定要一個motion計一次.

2007-12-19 19:06:00 補充:
樓上果個,mgh既h唔係向量,所以無負號架.


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