問題詳情：
如圖所示，質量為M且足夠長的木板在光滑的水平面上，其右端有一質量為m、可視為質點的滑塊，滑塊與木板間的動摩擦因數為μ。勁度係數為k的水平輕*簧的右端O固定不動，其自由端A與滑塊之間的距離為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，重力加速度大小為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*簧被壓縮的長度為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，設滑塊的最大速度為vm時木板的速度大小為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，所受的合外力為變力，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，合外力為零，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，合外力不為零，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
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