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A horizontal 52-n force is needed to slide a box across a flat surface at a constant velocity of 3.5 m/s. what is the coefficient of kinetic friction between the box and the floor?

Respuesta :

μk = 0.106 is the coefficient of kinetic friction between the box and the floor

When the box is moving across the flat surface with a constant velocity, the force applied to it must be equal to the kinetic frictional force applied to the object:

F = μk*W = μk*mg

Where,

F = Force applied on box = 52 N

m = mass of box = 50 kg

g = 9.8 m/s²

μk = coefficient of kinetic friction between box and floor

hence,

52 N  = μk(50 kg)(9.8 m/s²)

μk = (52 N)/(490 N)

μk = 0.106

Kinetic friction is the name for a force that exists between sliding parts. A force opposing a body's forward motion is applied to one that is moving on the surface. The kinetic friction coefficient between the two materials will determine the magnitude of the force.



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