Question

如圖所示，一個(gè)質(zhì)量為m=4kg的物體，受到與水平方向成37角斜向下方的推力F₁=20N的作用，在水平地面上移動(dòng)了距離l₁=2m后撤去推力，此后物體又滑行了l₂=1.4m，的距離后停止了運(yùn)動(dòng)．設(shè)物體與地面間的滑動(dòng)摩擦力為它們間彈力的0.2倍（sin37°=0.6                                                                     cos37°=0.8 g=10m/s²），求：                     

（1）推力F₁對(duì)物體做的功；                                                                                       

（2）全過(guò)程中摩擦力對(duì)物體所做的功；                                                                     

（3）推力作用時(shí)合外力對(duì)物體所做的總功．                                                              

                                                                                              

                                                                                                                                   

Answer 1

（1）根據(jù)恒力做功公式得：

W₁=F₁S₁cos37°=20×2×0.8J=32J

（2）由題意得

當(dāng)F₁作用時(shí)f₁=μ（mg+F₁ sin37°）=10.4N

撤去F₁時(shí)f₂=μmg=8N

則：W_f1=﹣f₁l₁=﹣10.4×2=﹣20.8J

W_f2=﹣f₂l₂=﹣8×1.4=﹣11.2J

所以W_f=W_f1+W_f2=﹣32J

（3）推力作用時(shí)合外力對(duì)物體所做的總功：W_總=W₁+W_f1=0J

答：（1）推力F₁對(duì)物體做的功為32J；

（2）全過(guò)程中摩擦力對(duì)物體所做的功為﹣32J

（3）推力作用時(shí)合外力對(duì)物體所做的總功為0J．