A 25 kg box is being pulled at a constant velocity with a tension force of 65 N. what is the coefficient of friction between the box and the ground?

The box is in equilibrium, so Newton's second law says

n + (-w) = 0

65 N + (-f ) = 0

where n denotes the magnitude of the normal force, w denotes the weight of the box, and f denotes the magnitude of the friction force.

The box has a weight of

w = (25 kg) (9.80 m/s²) = 245 N

so n = 245 N, too.

The friction force has magnitude

f = 65 N

and is proportional to the normal force by a factor of µ, the coefficient of kinetic friction. So we have

65 N = µ (245 N)   →  µ ≈ 0.26

the force on the thing and the distance the force moves the thing