如图所示，一足够长的水平传送带，以v＝2m/s的速度匀速顺时针转动运动．将一质量为m=1kg的物块无初速度地轻...

问题详情：

如图所示，一足够长的水平传送带，以v＝2 m/s的速度匀速顺时针转动运动．将一质量为m=1kg的物块无初速度地轻放在传送带左端，物块与传送带之间的动摩擦因数μ=0.05，

(1)求物块与传送带相对静止前物块受到的摩擦力Ff大小和方向(取g＝10 m/s2)

(2)若摩擦力使物块产生a1=0.5 m/s2的加速度，求物块与传送带共速所用时间t

(3)若关闭电动机让传送带以a2=1.5 m/s2的加速度减速运动，同时将该物块无初速度地放在传送带上，由于传送带比较光滑，物块在停止运动前始终无法与传送带保持相对静止且相对滑动过程中加速度的大小一直为a1=0.5 m/s2．求物块相对传送带滑动的位移大小L

【回答】

(1)f=μmg=0.5N······························································································· (1分)

方向向右·································································································· (1分)

(2) t0＝=4s·································································································· (2分)

(3设传送带减速到停止运动的距离x1

v2=2(-a2)x1         得x1＝m······································································ (2分)

设经t1物块与传送带速度恰好相等

a1t1= v- a2t1    得t1=1s············································································ (2分)

v共=0.5m/s································································································ (1分)

物块加速运动的距离x2＝t1=0.25 m····················································· (1分)

此后由题意知不能相对静止，则物体做加速度大小不变的匀减速运动，时间t2=t1.

减速运动的距离为x3＝t2=0.25 m························································ (1分)

相对滑动L＝x1-x2-x3＝m=0.83 m··························································· (1分)

知识点：牛顿运动定律的应用

题型：计算题