Explanation:
μ[tex]_{s}[/tex] = [tex]\frac{F_{appl} }{W}= \frac{F_{s} }{N}[/tex],
where, [tex]F_{appl}[/tex] is the horizontal pushing force,
W = mg is the weight of the crate directed downward,
[tex]F_{s}[/tex] is the static friction force-directed opposite to the horizontal pushing force and equal to it,
N is the force of reaction directed upward and equal to the weight of the crate.
From this formula we can find the horizontal pushing force required to just start the crate moving:
[tex]F_{appl} = F_{s} = u_{s}N = u_{s}mg[/tex]
= 0.760 [tex]\times[/tex] 60 kg [tex]\times[/tex] 9.8 m / s^2
= 447 N.
u[tex]_{k} = \frac{F_{appl} }{W} = \frac{F_{k} }{N}[/tex],
where, [tex]F_{appl}[/tex] is the horizontal pushing force,
W = mg is the weight of the crate directed downward,
[tex]F_{k}[/tex] is the kinetic friction force-directed opposite to the horizontal pushing force and equal to it,
N is the force of reaction directed upward and equal to the weight of the crate.
From this formula we can find the horizontal pushing force required to slide the crate across the dock at a constant speed:
[tex]F_{appl} = F_{k} = u_{k}N = u_{k}mg[/tex]
= 0.410 [tex]\times[/tex] 60 [tex]\times[/tex] 9.8
= 241 N.
