Question

如圖所示，質(zhì)量為M且足夠長的木板在光滑的水平面上，其右端有一質(zhì)量為m、可視為質(zhì)點的滑塊，滑塊與木板間的動摩擦因數(shù)為μ。勁度系數(shù)為k的水平輕彈簧的右端O固定不動，其自由端A與滑塊之間的距離為L�，F(xiàn)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(yùn)動，與彈簧接觸后經(jīng)過時間t，滑塊向右運(yùn)動的速度達(dá)到最大，設(shè)滑塊的速度始終小于木板的速度，彈簧的形變是在彈性限度內(nèi)，重力加速度大小為g，不計空氣阻力。求：

（1）滑塊剛接觸彈簧時滑塊的速度v₁大小和木板的速度v₂大�。�

（2）滑塊向右運(yùn)動的速度達(dá)到最大值的過程中，彈簧對滑塊所做的功W；

（3）滑塊向右運(yùn)動的速度最大值υ_m及其速度最大時滑塊與木板的右端之間的距離s。

Answer 1

【標(biāo)準(zhǔn)解答】（1）滑塊接觸彈簧之前，在滑動摩擦力作用下由靜止開始做勻加速直線運(yùn)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(tǒng)，由動量守恒定律：

Mv₀ = Mv₂ + mv₁································································································ ③（2分）

得：v₂ = v₀ –  ·················································································· ④（1分）

（2）滑塊接觸彈簧之后向右運(yùn)動的過程中，當(dāng)滑塊受到的滑動摩擦力與彈簧的彈力平衡時，滑塊的速度達(dá)到最大，此時彈簧被壓縮的長度為x，則：

μmg = kx

得x = ······································································································· ⑤（2分）

由于彈簧的彈力與彈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(yùn)動的速度達(dá)到最大值時，設(shè)滑塊的最大速度為v_m時木板的速度大小為v。在彈簧被壓縮的長度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 = mv_m² – mv₁²············································································· ⑧（2分）

 – μmg(L + x + s) = Mv² – Mv₀²··································································· ⑨（2分）

得

υ_m = ······················································································ ⑩（1分）

s = [v₀² – (v₀ –   – )²] –  – L··························· （11）（1分）

【思維點拔】本題的關(guān)鍵在于平時對滑塊與滑板疊加問題的積淀，對木板在滑動摩擦力作用做勻減速直線運(yùn)動，對滑塊在接觸彈簧之前做勻加速直線運(yùn)動，均可運(yùn)用牛頓運(yùn)動定律及運(yùn)動學(xué)公式計算，但對滑塊在接觸彈簧之后，所受的合外力為變力，則需用能量觀點解答。在接觸彈簧之前，對滑塊和木板組成的系統(tǒng)，合外力為零，所以也可以根據(jù)動量守恒定律來解答，但在接觸彈簧之后，對滑塊和木板組成的系統(tǒng)，合外力不為零，所以不能根據(jù)動量守恒定律來解答。對于彈簧彈力做功問題，由于彈簧的彈力與彈簧被壓縮的長度成線性關(guān)系，所以可以用平均力來計算彈力做功。