Question

如圖所示,勁度系數(shù)為k=200 N/m的輕彈簧一端固體在墻上,另一端連接一質(zhì)量為M=8 kg的小車a,開(kāi)始時(shí)小車靜止,其左端位于O點(diǎn),彈簧沒(méi)有發(fā)生形變.質(zhì)量為m=1 kg的小物塊b靜止于小車的左側(cè),距O點(diǎn)s=3 m,小車與水平面間的摩擦不計(jì),小物塊與水平面間的動(dòng)摩擦因數(shù)為μ=0.2,取g=10 m/s².今對(duì)小物塊施加大小為F=8 N的水平恒力使之向右運(yùn)動(dòng),并在與小車發(fā)生碰撞前的瞬間撤去該力,碰撞后小車做振幅為A=?0.2 m的簡(jiǎn)諧運(yùn)動(dòng).已知小車做簡(jiǎn)諧運(yùn)動(dòng)周期公式為T(mén)=2π,彈簧的彈性勢(shì)能公式為E_p=kx²(x為彈簧的形變量),則:

(1)小物塊與小車碰撞前瞬間的速度是多大?

(2)小車做簡(jiǎn)諧運(yùn)動(dòng)過(guò)程中彈簧最大彈性勢(shì)能是多少?小車的最大速度為多大?

(3)小物塊最終停在距O點(diǎn)多遠(yuǎn)處?當(dāng)小物塊剛停下時(shí)小車左端運(yùn)動(dòng)到O點(diǎn)的哪一側(cè)?

Answer 1

解析:(1)設(shè)碰撞前瞬間,小物塊b的速度為v₁

小物塊從靜止開(kāi)始運(yùn)動(dòng)到剛要與小車發(fā)生碰撞的過(guò)程中,根據(jù)動(dòng)能定理可知

Fs-μmgs=mv₁²                                                              ①

解得v₁=6 m/s.                                                                 ②

(2)由于小車簡(jiǎn)諧運(yùn)動(dòng)的振幅是0.2 m,所以彈簧的最大形變量為x=A=0.2 m

根據(jù)彈性勢(shì)能的表達(dá)式可知最大彈性勢(shì)能E_pm=kA²                               ③

解得E_pm=4 J                                                                  ④

根據(jù)機(jī)械能守恒定律可知小車的最大動(dòng)能應(yīng)等于彈簧的最大彈性勢(shì)能,

所以kA²=Mv_m²                                                           ⑤

解得小車的最大速度v_m=1 m/s.                                                  ⑥

(3)小物塊b與小車a碰撞后,小車a的速度為v_m,設(shè)此時(shí)小物塊的速度為v₁′,設(shè)向右為正方向,由動(dòng)量守恒定律有mv₁=mv₁′+Mv_m                                               ⑦

解得v₁′=-2 m/s                                                                ⑧

接著小物塊b向左做勻減速運(yùn)動(dòng)一直到停止,設(shè)位移是s₁,所經(jīng)歷的時(shí)間為t₁,根據(jù)動(dòng)能定理可知-μmgs₁=0-mv₁′²                                                           ⑨

解得s₁=1 m                                                                   ⑩

物塊做減速運(yùn)動(dòng)時(shí)的加速度為a==μg=2 m/s^{2                                                  11}

t₁==1 s                                                               12

小車a振動(dòng)的周期T=2π≈1.26 s                                       13

由于T>t₁>T,所以小車a在小物塊停止時(shí)在O點(diǎn)的左側(cè)，并向右運(yùn)動(dòng).              14