問題詳情：
如圖所示，一足夠長的水平傳送帶，以v＝2 m/s的速度勻速順時針轉動運動．將一質量爲m=1kg的物塊無初速度地輕放在傳送帶左端，物塊與傳送帶之間的動摩擦因數μ=0.05，
(1)求物塊與傳送帶相對靜止前物塊受到的摩擦力Ff大小和方向(取g＝10 m/s2)
(2)若摩擦力使物塊產生a1=0.5 m/s2的加速度，求物塊與傳送帶共速所用時間t
(3)若關閉電動機讓傳送帶以a2=1.5 m/s2的加速度減速運動，同時將該物塊無初速度地放在傳送帶上，由於傳送帶比較光滑，物塊在停止運動前始終無法與傳送帶保持相對靜止且相對滑動過程中加速度的大小一直爲a1=0.5 m/s2．求物塊相對傳送帶滑動的位移大小L
【回答】
(1)f=μmg=0.5N······························································································· (1分)
方向向右·································································································· (1分)
(2) t0＝
=4s·································································································· (2分)
(3設傳送帶減速到停止運動的距離x1
v2=2(-a2)x1 得x1＝
m······································································ (2分)
設經t1物塊與傳送帶速度恰好相等
a1t1= v- a2t1 得t1=1s············································································ (2分)
v共=0.5m/s································································································ (1分)
物塊加速運動的距離x2＝
t1=0.25 m····················································· (1分)
此後由題意知不能相對靜止，則物體做加速度大小不變的勻減速運動，時間t2=t1.
減速運動的距離爲x3＝
t2=0.25 m························································ (1分)
相對滑動L＝x1-x2-x3＝
m=0.83 m··························································· (1分)
知識點：牛頓運動定律的應用
題型：計算題
