问题详情:
如图所示,质量为M且足够长的木板在光滑的水平面上,其右端有一质量为m、可视为质点的滑块,滑块与木板间的动摩擦因数为μ。劲度系数为k的水平轻*簧的右端O固定不动,其自由端A与滑块之间的距离为L。现给木板以水平向右的瞬时速度v0,滑块将由静止开始向右运动,与*簧接触后经过时间t,滑块向右运动的速度达到最大,设滑块的速度始终小于木板的速度,*簧的形变是在**限度内,重力加速度大小为g,不计空气阻力。求:
(1)滑块刚接触*簧时滑块的速度v1大小和木板的速度v2大小;
(2)滑块向右运动的速度达到最大值的过程中,*簧对滑块所做的功W;
(3)滑块向右运动的速度最大值υm及其速度最大时滑块与木板的右端之间的距离s。
【回答】
【标准解答】(1)滑块接触*簧之前,在滑动摩擦力作用下由静止开始做匀加速直线运动了L距离,由动能定理:
μmgL=mv12·································································································· ①(2分)
得:v1 =······························································································ ②(1分)
对滑块和木板组成的系统,由动量守恒定律:
Mv0 =Mv2 + mv1································································································ ③(2分)
得:v2 =v0 – ·················································································· ④(1分)
(2)滑块接触*簧之后向右运动的过程中,当滑块受到的滑动摩擦力与*簧的*力平衡时,滑块的速度达到最大,此时*簧被压缩的长度为x,则:
μmg=kx
得x=······································································································· ⑤(2分)
由于*簧的*力与*簧被压缩的长度成正比,所以有
W=–kx · x=–················································································ ⑥(2分)
(3)滑块向右运动的速度达到最大值时,设滑块的最大速度为vm时木板的速度大小为v。在*簧被压缩的长度x的过程中,对木板由动量定理:
–μmgt=Mv–Mv2····························································································· ⑦(2分)
得v=v0 – –
分别对滑块和木板,由动能定理:
μmgx + W=mvm2 – mv12············································································· ⑧(2分)
–μmg(L + x + s) =Mv2 –Mv02··································································· ⑨(2分)
得
υm =······················································································ ⑩(1分)
s=[v02 – (v0 – –)2] – –L··························· (11)(1分)
【思维点拔】本题的关键在于平时对滑块与滑板叠加问题的积淀,对木板在滑动摩擦力作用做匀减速直线运动,对滑块在接触*簧之前做匀加速直线运动,均可运用牛顿运动定律及运动学公式计算,但对滑块在接触*簧之后,所受的合外力为变力,则需用能量观点解答。在接触*簧之前,对滑块和木板组成的系统,合外力为零,所以也可以根据动量守恒定律来解答,但在接触*簧之后,对滑块和木板组成的系统,合外力不为零,所以不能根据动量守恒定律来解答。对于*簧*力做功问题,由于*簧的*力与*簧被压缩的长度成线*关系,所以可以用平均力来计算*力做功。
知识点:专题五 动量与能量
题型:综合题