更新时间2022-01-18 21:51:48
在线等!!!
摩檫力 f = μm2g = 0.2 * 10/2 = 1N;
m2减速度 a = f/m2 = 1/0.2 = 5 m/s^2;
v = at = 5t;
1、由动量定理,ft = m2v0 - ( m1 + m2 )v t = 0.2 * 2 - 0.5 * 5t
t = 4/35 s ;
2、v = 5 * 4/35 = 4/7 m/s
S = ( v0'^2 - v^2 )/(2a) < 1.5m
v0'^2 < 2 * 5 * 1.5 + (4/7)^2
v0' < √751/7 ≈ 3.9 m/s 。
1、设到达B处速度为 v;
v = at = 2a;
2sa = v^2,40a = 4a^2 a = 10;
f = 30 - μmg = ma
μ = ( 30 - ma )/mg = 30/20 - 1 = 0.5;
2、设B点速度为 v;
水平拉力 F1 = 30cos37° = 24N
竖直拉力 F2 = 30sin37° = 18N
摩檫力 f = μ( mg - F2 ) = 0.5 * ( 2 * 10 - 18 ) = 1N
由动量定理,( F1 - f )t = mv = m * 2s/t
t^2 = 4s/( F1 - f ) = 80/23
t = √( 80/23 ) ≈ 1.865 s 。
物块受到的合外力为摩檫力;
f = μmg = 4 * 10/2 = 20N;
1、物块相对传送带v0是 -20m/s;
由动能定理,fs = m( vc^2 - v0'^2 )/2
s = 0 - (-20)^2/10 = -40m;
2、物块相对传送带v0是 40 - 20 = 20m/s;
s = 20^2/10 - 0 = 40m;
3、物块相对传送带v0是 -40 - 20 = -60m/s;
s = 0 - (-60)^2/10 = 360m 。