Answer:
[tex]\mu_{k}=0.12[/tex]
Explanation:
In order to solve this problem we must first start by drawing a diagram of the situation. (See attached picture).
So we can start solving this problem by doing a sum of the forces in the y-direction so we get:
[tex]\sum F_{y}=0[/tex]
which yields:
[tex]N+F_{y}-W=0[/tex]
which can be solved for N so we get:
[tex]N=W-F_{y}[/tex]
we know that:
[tex]F_{y}=Fsin\theta[/tex]
and that
W=mg
so we can substitute that into our original equation so we get:
[tex]N=mg-Fsin\theta[/tex]
we can also substitute the provided data here so we get:
[tex]N=(60kg)(9.81m/s^{2})-75Nsin(30^{o})[/tex]
which yields:
N=551.1N
Next we can do a sum of forces in the x-direction so we get:
[tex]\sum F_{x}=0[/tex]
which gives us:
[tex]F_{x}-f=0[/tex]
We know that the friction is given by the following formula:
[tex]f=N\mu_{k}[/tex]
and that:
[tex]F_{x}=Fcos \theta[/tex]
so we can substitute them into our sum of forces so we get:
[tex]Fcos \theta - N\mu_{k}=0[/tex]
we can now solve this for [tex]\mu_{k}[/tex]:
which yields:
[tex]\mu_{k}=\frac{Fcos \theta}{N}[/tex]
now we can substitute the provided data so we get:
[tex]\mu_{k}=\frac{(75N)cos 30^{o}}{551.1N}=0.12[/tex]
So the coefficient of kinetic friction between the sled and the snow is 0.12.
