一个质量mA为4kg的木板A放在水平面C上,木板与水平面间的动摩擦因数u=0.24,木板右上端放着质量m B为1.0k的小物块B(视为质点),它们均处于静止状态,木板A 突然受到水平向右的瞬时冲量I=12N.s

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一个质量mA为4kg的木板A放在水平面C上,木板与水平面间的动摩擦因数u=0.24,木板右上端放着质量m B为1.0k的小物块B(视为质点),它们均处于静止状态,木板A 突然受到水平向右的瞬时冲量I=12N.s
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一个质量mA为4kg的木板A放在水平面C上,木板与水平面间的动摩擦因数u=0.24,木板右上端放着质量m B为1.0k的小物块B(视为质点),它们均处于静止状态,木板A 突然受到水平向右的瞬时冲量I=12N.s
一个质量mA为4kg的木板A放在水平面C上,木板与水平面间的动摩擦因数u=0.24,木板右上端放着质量m B为1.0k的小物块B(视为质点),它们均处于静止状态,木板A 突然受到水平向右的瞬时冲量I=12N.s作用后开始运动.当小物块B滑落木板时,木板的动能EKA=8.0J,小物块的动能EKB=0.50J,重力加速度取10m/s2.求:
1)瞬时冲量作用结束时木板的速度v0;
2)木板的长度L

一个质量mA为4kg的木板A放在水平面C上,木板与水平面间的动摩擦因数u=0.24,木板右上端放着质量m B为1.0k的小物块B(视为质点),它们均处于静止状态,木板A 突然受到水平向右的瞬时冲量I=12N.s
1)I=mB*V0
V0=I/mB=12 / 1=12 m/s
2)对B,F*SB=EkB
F* t =mB*VB
对A,[ F+u*(mA+mB*)g ]*SA=(mA*V0^2 / 2)-EkA
[ F+u*(mA+mB*)g ]* t =mA*(V0-VA)
SA-SB=L
以上五式联立,得
L=21.3125米  文字输入有限.

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如图所示,质量mA为4kg的木板A放在水平面间C上,木板和水平面间的动摩擦因数为0.2一个质量mA为4kg的木板A放在水平面C上,木板与水平面间的动摩擦因数u=0.24,木板右上端放着质量m B为1.0k的小物块 ) 质量mA为4.0kg的木板A放在水平面C上,质量mA为4.0kg的木板A放在水平面C上,木板与水平面间的动摩擦因数μ为0.24,木板右端放着质量mB为1.0kg的小物块B(视为质点),它们均处于静止状态.木块突 一个质量mA为4kg的木板A放在水平面C上,木板与水平面间的动摩擦因数u=0.24,木板右上端放着质量m B为1.0k的小物块B(视为质点),它们均处于静止状态,木板A 突然受到水平向右的瞬时冲量I=12N.s ,质量mA为4.0kg的木板A放在水平面C上,木板与水平面间的动摩擦因数μ为0.24,木板右端放着质量mB为1.0kg的小物块B(视为质点),它们均处于静止状态.木板突然受到水平向右的12N•s的瞬时冲量I 物体 A 放在足够长的木板 B 上,木板 B 静止于水平面.已知 A 的质量 mA 和 B 的质量 mB 均为 2.0kg,A、B物体 A 放在足够长的木板 B 上,木板 B 静止于水平面.已知 A 的质量 mA 和 B 的质量 mB 均为 2.0kg,A 滑块与木板间的摩擦生热问题大侠,我想请教一道题力学题,希望得到解答.题如下:质量Ma 为4.0kg的木板A放在水平面C上,木板与平面间的动摩擦因素为0.24,木板右端放着质量Mb为1kg的小物块B(视 不难 但要完整过程和正确答案质量mA=2.0kg的木板A放在水平面C上,A与C之间的动摩擦因数μ=0.2.木板A右端放着质量m=1.0 kg的小木块B(可视为质点),都处于静止状态.现给木板A一个向右的瞬时冲量 质量为Mb=14kg的木板B放在水平面上.质量为Ma=10kg的木箱A放在木板B上一根轻绳栓在左边的一个木桩上,另一端栓在A上绳子绷紧是于水平面成37度.A与B之间的摩擦因数为0.5.B与水平面的摩擦因数为0 质量为Mb=14kg的木板B放在水平面上.质量为Ma=10kg的木箱A放在木板B上,一根轻绳栓在左边的一个木桩上,另一端栓在A上绳子绷紧是于水平面成37度.A与B之间的摩擦因数为0.5.B与水平面的摩擦因数为 一道动能、动量的物理题::,如图所示,质量为4kg的木板A在光滑水平面C上,右端放着1kg的物块,刚开始均静止,A与C、A与B之间的动摩擦系数均为0.2,给木板一个 V1= 3 m/s的速度开始运动,当物块离 如图所示,质量为mB=14kg的木板B放在水平地面上,质量为mA=10kg的木箱A在木板B上.一根轻绳一端拴在木箱上,另一端拴在地面的木桩上,绳绷紧时与水平面的夹角为=37°.已知木箱A与木板B之间的动摩 质量Ma为4千克的 木板放在水平面C上,木板遇水平面间的 动摩因素为0.24,木板右端放着质量Mb为1千克的 小物体B,他们均处于静止状态.木板突然受到水平向右的 12N.S的 瞬时冲量I作用开始运动,当 质量Ma为4千克的 木板放在水平面C上,木板遇水平面间的 动摩因素为0.24,木板右端放着质量Mb为1千克的 小物体B,他们均处于静止状态.木板突然受到水平向右的 12N.S的 瞬时冲量I作用开始运动,当 牛顿定律综合运用物体 A 放在足够长的木板 B 上,木板 B 静止于水平面.已知 A 的质量 mA 和 B 的质量 mB 均为 2.0kg,A、B 之间的动摩擦因数 1=0.2,B 与水平面之间的动摩 擦因数 2=0.1,最大静摩擦力与 牛顿第二定律高中物理题物体A放在足够长的木板B上,木板B静止于水平面.已知A的质量mA和B的质量mB均为2.0kg,AB之间的动摩擦因数=0.2,B与水平面之间的动摩擦因数=0.1.最大静摩擦力与滑动摩擦力 如图所示,物体A放在足够长的木板B上,木板B静止于水平面.t=0时,电动机通过水平细绳以恒力F拉木板B,使它做初速度为零,加速度aB=1.0m/s²的匀加速直线运动.已知A的质量mA和B的质量mg均为2.0kg,A 如图所示,物体A放在足够长的木板B上,木板B静止于水平面上.已知A的质量为ma和B的质量为mb均为2.0kg,A、B之间的动摩擦因数为0.2,B于水平面之间的动摩擦因数为0.1,最大静摩擦力与滑动摩擦力大 如图所示,质量为4kg的木块放在木板上.如图所示,质量为4kg的木块放在木板上,当木板与水平面的夹角为37°时,木块恰能沿木板匀速下滑.求:(1)求木块与木板间的动摩擦因数(2)若将木板水