Answer:[tex]F_{v} =\mu_{k} mg[/tex]
Magnitude of the force is 2601.9 N
Explanation:
m = 450 kg
coefficient of static friction μs = 0.73
coefficient of kinetic friction is μk = 0.59
The force required to start crate moving is [tex]F_{s} =\mu_{s} mg[/tex].
but once crate starts moving the force of friction is reduced [tex]F_{v} =\mu_{k} mg[/tex].
Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie [tex]F_{v} =\mu_{k} mg[/tex].
Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.
Magnitude of the force
[tex]F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N[/tex]