Question

如圖5-1-3所示，質(zhì)量m_A為4.0 kg的木板A放在水平面C上，木板與水平面間的動(dòng)摩擦因數(shù)μ為0.24,木板右端放著質(zhì)量m_B為1.0 kg的小物塊B（視為質(zhì)點(diǎn)），它們均處于靜止?fàn)顟B(tài).木板突然受到水平向右的12 N·s的瞬時(shí)沖量I作用開始運(yùn)動(dòng)，當(dāng)小物塊滑離木板時(shí)，木板的動(dòng)能E_kA為8.0 J，小物塊的動(dòng)能E_kB為0.50 J，重力加速度取10 m/s².求：

圖5-1-3

（1）瞬時(shí)沖量作用結(jié)束時(shí)木板的速度v₀;

(2)木板的長度L.

Answer 1

（1）3.0 m/s  （2）0.50 m

解析:

(1)設(shè)水平向右為正方向，有I=m_Av₀                                     ①

代入數(shù)據(jù)解得v₀=3.0 m/s                                                     ②

（2）設(shè)A對(duì)B、B對(duì)A、C對(duì)A的滑動(dòng)摩擦力的大小分別為F_AB、F_BA和F_CA，B在A上滑行的時(shí)間為t，B離開A時(shí)A和B的速度分別為v_A和v_B，分別對(duì)A、B應(yīng)用動(dòng)量定理,有

-（F_BA+F_CA）t=m_Av_A-m_Av₀                                                    ③

F_ABt=m_Bv_B                                                                  ④

其中F_AB=F_BA

F_CA=μ(m_A+m_B)g                                                            ⑤

設(shè)A、B相對(duì)于C的位移大小分別為s_A和s_B，分別對(duì)A、B應(yīng)用動(dòng)能定理,有

-（F_BA+F_CA）s_A=m_Av_A²-m_Av₀²                                             ⑥

F_ABs_B=E_Kb                                                                  ⑦

動(dòng)量與動(dòng)能之間的關(guān)系為

m_Av_A=                                                            ⑧

m_Bv_B=                                                             ⑨

木板A的長度L=s_A-s_B                                                        ⑩

代入數(shù)據(jù)解得L=0.50 m.