a 4kg body is placed on horizontal ground.The coefficient of dynamic friction between the body and the ground is 0,2.The object is acted upon by a force of 30 N that encloses an angle of 20 with horizontal.Find the friction force and the acceleration

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a 4kg body is placed on horizontal ground.The coefficient of dynamic friction between the body and the ground is 0,2.The object is acted upon by a force of 30 N that encloses an angle of 20 with horizontal.Find the friction force and the acceleration
a 4kg body is placed on horizontal ground.The coefficient of dynamic friction between the body and the ground is 0,2.The object is acted upon by a force of 30 N that encloses an angle of 20 with horizontal.Find the friction force and the acceleration of the body

a 4kg body is placed on horizontal ground.The coefficient of dynamic friction between the body and the ground is 0,2.The object is acted upon by a force of 30 N that encloses an angle of 20 with horizontal.Find the friction force and the acceleration
frictional force
0.2*(4*9.8-30*Sin20)
= 5.8 N
acceleration
(30*Cos20-0.2*(4*9.8-30*Sin20))/4
= 5.6 m/s^2

那些公式早忘光了

我看你问了好几次了
其实你可以这样
先找人翻译再找人做
因为一般懂英语的逻辑思维都不行
逻辑思维好的一般都不大懂英语

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翻译:一个4kg的物体放置在水平面上.物体和地面之间的摩擦因数是0.2.物体受一个与水平面夹角是20度的力的作用,力大小是30N.求摩擦力和物体的加速度.
解答:竖直方向上物体没有加速度,受力平衡:
Fsin20度+N=mg 其中,F是拉力,N是支持力,令g=9.8m/s^2
f=μN f是摩擦力, 解出:f=μ(mg-Fsin20度)=5.83N
水平方向上有加速度a.则根据牛顿第二定律:Fcos20度-f=ma
解出:a=5.59m/s^2
In the vertical direction, the body doesn't have acceleration:
Fsin20+N=mg,
F is the force that act upon the body, N is the force of support from the ground. Let g be g=9.8m/s^2
f=μN,
f is the friction force, μ is the coefficient of dynamic friction between the body and the ground.
So, f=5.83N
In the horizontal direction, according to Newton's Second Law: (Fcos20-f)=ma
a is the acceleration.
So, a=5.59m/s^2

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