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
