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Sagot :
15.9 N
Explanation:
Let's assume that the downward direction on the inclined is the (+)-direction. Since the block is in equilibrium, the x-component of its weight is pointing in the +x-direction and the frictional force [tex]f_s[/tex] is pointing up the incline. So the net force acting parallel to the incline can be written as
[tex]mg\sin36 - f_s = 0 \Rightarrow \mu N = mg\sin36[/tex]
where N is the normal force.
The net force perpendicular to the incline can be written as
[tex]N - mg\cos36 = 0 \Rightarrow N = mg\cos36[/tex]
or
[tex]N = (2\:\text{kg})(9.8\:\text{m/s}^2)\cos36 = 15.9\:\text{Newtons}[/tex]
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