問題詳情:
如圖所示,質量爲M且足夠長的木板在光滑的水平面上,其右端有一質量爲m、可視爲質點的滑塊,滑塊與木板間的動摩擦因數爲μ。勁度係數爲k的水平輕*簧的右端O固定不動,其自由端A與滑塊之間的距離爲L。現給木板以水平向右的瞬時速度v0,滑塊將由靜止開始向右運動,與*簧接觸後經過時間t,滑塊向右運動的速度達到最大,設滑塊的速度始終小於木板的速度,*簧的形變是在**限度內,重力加速度大小爲g,不計空氣阻力。求:
(1)滑塊剛接觸*簧時滑塊的速度v1大小和木板的速度v2大小;
(2)滑塊向右運動的速度達到最大值的過程中,*簧對滑塊所做的功W;
(3)滑塊向右運動的速度最大值υm及其速度最大時滑塊與木板的右端之間的距離s。
【回答】
【標準解答】(1)滑塊接觸*簧之前,在滑動摩擦力作用下由靜止開始做勻加速直線運動了L距離,由動能定理:
μmgL=mv12·································································································· ①(2分)
得:v1 =······························································································ ②(1分)
對滑塊和木板組成的系統,由動量守恆定律:
Mv0 =Mv2 + mv1································································································ ③(2分)
得:v2 =v0 – ·················································································· ④(1分)
(2)滑塊接觸*簧之後向右運動的過程中,當滑塊受到的滑動摩擦力與*簧的*力平衡時,滑塊的速度達到最大,此時*簧被壓縮的長度爲x,則:
μmg=kx
得x=······································································································· ⑤(2分)
由於*簧的*力與*簧被壓縮的長度成正比,所以有
W=–kx · x=–················································································ ⑥(2分)
(3)滑塊向右運動的速度達到最大值時,設滑塊的最大速度爲vm時木板的速度大小爲v。在*簧被壓縮的長度x的過程中,對木板由動量定理:
–μmgt=Mv–Mv2····························································································· ⑦(2分)
得v=v0 – –
分別對滑塊和木板,由動能定理:
μmgx + W=mvm2 – mv12············································································· ⑧(2分)
–μmg(L + x + s) =Mv2 –Mv02··································································· ⑨(2分)
得
υm =······················································································ ⑩(1分)
s=[v02 – (v0 – –)2] – –L··························· (11)(1分)
【思維點拔】本題的關鍵在於平時對滑塊與滑板疊加問題的積澱,對木板在滑動摩擦力作用做勻減速直線運動,對滑塊在接觸*簧之前做勻加速直線運動,均可運用牛頓運動定律及運動學公式計算,但對滑塊在接觸*簧之後,所受的合外力爲變力,則需用能量觀點解答。在接觸*簧之前,對滑塊和木板組成的系統,合外力爲零,所以也可以根據動量守恆定律來解答,但在接觸*簧之後,對滑塊和木板組成的系統,合外力不爲零,所以不能根據動量守恆定律來解答。對於*簧*力做功問題,由於*簧的*力與*簧被壓縮的長度成線*關係,所以可以用平均力來計算*力做功。
知識點:專題五 動量與能量
題型:綜合題