A 550N object has a coefficient of .012 against a smooth surface. What force is needed to keep the object sliding?(ANSWER to 1 decimal place, no
units)

Vertically, the object is in equilibrium, so that the net force in this direction is

∑ F (vertical) = n - mg = 0

where n is the magnitude of the normal force due to the contact between the object and surface. You're given that the object's weight is mg = 550 N, so n = 550 N as well.

Horizontally, the net force would be

∑ F (horizontal) = p - f = 0

where p is the magnitude of the applied force and f is the magnitude of (kinetic) friction opposing p. Now,

f = 0.012n = 0.012 (550 N) = 6.6 N

so that you need to apply a force of p = 6.6 N to keep the object sliding at a steady pace.

ill do it for 100, also this isnt 80 its 40