DE for modelling motion of a box

[acrusera]

So if you were to slide a box-like object down a straight ramp with a single gradient (its literally just an incline no curve). And you attempt to model it accurately would this be correct?
Given the DE
ma=mg-kv
where a=dv/dt, mg=force of gravity, kv= force of friction

Anyway my main question, is k the coefficient of kinetic friction? Does k=μ?

 

[Suraj M]

What is µ according to you? Coefficient of static friction? then no ,k≠ µ

 

[PeroK]

So if you were to slide a box-like object down a straight ramp with a single gradient (its literally just an incline no curve). And you attempt to model it accurately would this be correct?
Given the DE
ma=mg-kv
where a=dv/dt, mg=force of gravity, kv= force of friction

Anyway my main question, is k the coefficient of kinetic friction? Does k=μ?

I think you've got your equations mixed up.

$ma = mg - kv$

Is for a body falling under gravity with air resistance proportional to its velocity.

 

[Suraj M]

Is for a body falling under gravity with air resistance proportional to its velocity.

k=$6 \pi \eta r$ ??
then what about the ramp?

 

[HallsofIvy]

The "coefficient of friction", $\mu$ is the number that multiplies the normal force on an object to give the friction force. An object of mass m, has weight mg. On a ramp making angle $\theta$ with the horizontal, we can divide the weight into components parallel to and normal to the ramp. The component normal to the ramp is $mg cos(\theta)$.