※ 引述《alwyner (Time is money!!)》之銘言： : ※ [本文轉錄自 Physics 看板] : 作者: alwyner (Time is money!!) 看板: Physics : 標題: [題目] 牛頓力學... : 時間: Sun Apr 27 13:34:07 2008 : 1.阻力R = kv ，質量m的小鋼球從油的表面(y=0)靜止釋放，k為常數，導出在h深處，此 : 球所到達的速度v。 : Ans: h =? y is negative for downward v is negative for downward v = y' my" = -mg - ky' set v = y' => v' = -g - (k/m)v => v = Aexp(-kt/m) - mg/k v(t = 0) = 0 => A = mg/k And y' = v = (mg/k)[exp(-kt/m) - 1] 2 => y = -g(m/k) exp(-kt/m) - mgt/k + C 2 y(t = 0) = 0 => C = g(m/k) Therefore, 2 2 y = -g(m/k) exp(-kt/m) - mgt/k + g(m/k) v = (mg/k)[exp(-kt/m) - 1] The question asks us to find the │v│ given h, but your answer is h = ?? I'm confused. : 2. ／ : m ／＼／ │ : ＼／ │ : 夾角 θ ／ M │ : ↘ ／ │ : ／ │ : │￣￣ │ : ￣￣￣￣￣￣￣￣￣ : 忽略所有摩擦，求a)M的加速度 b)m相對於M的加速度。 Let A(->) be the acceleration of M, a the component of the acceleration of m "relative" to M in the x-direction . Let N' be the normal force between m and M, N between M and floor. positive direction:upward and rightward Assume A>0:(->), a>0:(<-) Momentum conservation in the x-directon plus differentiation about time: => MA + m(-a + A) = 0 ---[1] Force applied in the y-direction: => -ma*tanθ = -(m+M)g + N ---[2] For big block in the y-direction: => -N'cosθ - Mg + N = 0 ---[3] For big block in the x-direction: => N'sinθ = MA ---[4] There is enough information for you to solve A. ---------------------------- From [3] and [4] => N = M(A*cotθ+ g ) ---[5] 2 Substitute [1] and [5] into [2] => A = mgsinθcosθ/(M + m*sinθ) the acceleration of m relative to M is a*secθ 2 From [1] => a*secθ = (M + m)A*secθ/m = (M + m)g*sinθ/(M + m*sinθ) Period. --

※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 122.124.100.190 ※ 編輯: Honor1984 來自: 122.124.100.190 (05/13 03:13)