[問題] 一題功與能

看板Physics作者 (阿信)時間14年前 (2011/07/06 13:30), 編輯推噓2(2推 0噓 13→)
留言15則, 4人參與, 最新討論串1/1
An object of mass m is initially at rest on a frictionless horizontal plane. If a constant power P acts on this object for t duration, how far will it travel during this interval? 我自己的想法是 作功 W = Pt = Fx = d(mv)/dt·x ==> mx(dv) = Pt(dt) ==> mxv = (1/2)Pt^2 ==> mx(dx/dt) = (1/2)Pt^2 ==> mx(dx) = (1/2)Pt^2(dt) ==> (1/2)mx^2 = (1/6)Pt^3 ==> x = [(Pt^3)/3m]^(1/2) 要考試了 不知道自己的觀念有沒有正確 所以PO上來 有錯請大力鞭@@ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.37.169.218
推
07/06 13:42, , 1F
第一行就錯了...因為F不是定值
07/06 13:42, 1F
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07/06 13:42, , 2F
所以W != Fx
07/06 13:42, 2F
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07/06 13:47, , 3F
然後第二行到第三行也錯了...如果你認為x和v有關係(如
07/06 13:47, 3F
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07/06 13:47, , 4F
第四行所指)，那在這行就不能這麼簡單的積分
07/06 13:47, 4F
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07/06 13:49, , 5F
建議原PO想一下，要用來標定系統狀態的參數要用x還是t
07/06 13:49, 5F
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07/06 13:50, , 6F
我的解法是：P=Fv=const => F=P/v=ma
07/06 13:50, 6F
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07/06 13:51, , 7F
應該是用t吧?
07/06 13:51, 7F
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07/06 13:52, , 8F
(P/m)dt=vdv => Pt/m = (v^2)/2 + C
07/06 13:52, 8F
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07/06 13:53, , 9F
init rest => C=0
07/06 13:53, 9F
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07/06 13:53, , 10F
剩下再做一次積分就結束了
07/06 13:53, 10F
推
07/06 13:57, , 11F
老題目了啦
07/06 13:57, 11F
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07/06 13:58, , 12F
感謝@@ 了解了
07/06 13:58, 12F
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07/06 13:58, , 13F
我記得是Halliday功能定理例題最後一題
07/06 13:58, 13F
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08/13 16:22, , 14F
init rest = https://muxiv.com
08/13 16:22, 14F
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09/17 14:20, , 15F
init rest = https://daxiv.com
09/17 14:20, 15F
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文章代碼(AID): #1E4_CA_3 (Physics)