Question

如圖14所示，質(zhì)量為M的小車放在光滑水平面上，小車上用細(xì)線懸掛一質(zhì)量為m的小球，M＞m.用一力F水平向右拉小球，使小球和車一起以加速度a向右運(yùn)動(dòng)，細(xì)線與豎直方向成α角，細(xì)線的拉力為F₁.若用一力F′水平向左拉小車，使小球和車一起以加速度a′向左運(yùn)動(dòng)時(shí)，細(xì)線與豎直方向也成α角，細(xì)線的拉力為F₁′，則（　　）?

A.a′=a,F₁′=F₁

B.a′＞a,F₁′=F₁?

C.a′＜a,F₁′=F₁

D.a′＞a,F₁′＞F₁?

圖14

Answer 1

B?

解析：本題以常見(jiàn)的小車模型為背景，考查了整體法和隔離法在牛頓第二定律中的應(yīng)用.對(duì)下圖整體分析，可得

F=(M+m)a                                                                                         ①?

隔離m分析，如圖所示，F(xiàn)_合=F-F₁sinα=ma                                         ②

F_合=0，F(xiàn)₁cosα=mg                                                                             ③

由③式得F₁=mg/cosα代入②式得F=mgtanα+ma代入①式得mgtanα+ma=(M+m)a解得a=mgtanα.對(duì)上圖進(jìn)行整體分析F′=(M+m)a                                 ′④

隔離m進(jìn)行受力分析?F₁′cosα=mg                                                    ⑤?

F₁′=mg/cosα，F(xiàn)₁′sinα=ma′                                                                   ⑥?

由⑤⑥式得a′=gtanα.故F₁=F₁′,a′＞a,故B正確.