The normal force that the ground exerts on the sled will be given as
[tex]F_{N} =459.01 N[/tex]
Here we have the following details
Weight of the sled = 485 N
coefficient of kinetic friction between the sled and the ground = 0.200
The rope is at an angle of above the horizontal = 12
pull on the rope with a force = 125N
Now applying newtons law
[tex]F_{N} +F_{Y} -W=0[/tex]
Now from the Pythagorean theorem, we can calculate the component of the force at Y-axis
[tex]Sin(12)=\dfrac{F_{y} }{F}[/tex]
[tex]F_{y}=FSin(12)\dfrac{ }{}[/tex]
Now putting this value
[tex]F_{N}=W-FSin12[/tex]
[tex]F_{N}=485-(125)Sin12[/tex]
[tex]F_{N}=459.01N[/tex]
The normal force that the ground exerts on the sled will be given as
[tex]F_{N} =459.01 N[/tex]
To know more about the Resolution of the Force follow
https://brainly.com/question/1858958
