DC motor question

[theone]

Homework Statement

A machine, with an armature resistance of 0.16 ohms and K$\phi$ is to drive a load that requires a torque that is proportional to its speed. One point on the mechanical torque-speed relation is 16 Nm at 400 r/min. If the armature terminal voltage is 50 V, at what steady-state speed will the load be driven?

Homework Equations

[/B]
$T=K\phi i_a$, for a constant machine speed

The torque speed relation is
$w = \frac{V_t}{K\phi} - \frac{R_a}{(K\phi)^2}T$

The Attempt at a Solution

the only thing missing for me to use the torque speed relation is the load torque, but I don't understand what they did here to get it. I don't know what A is.

 

[andrevdh]

The problem states that the torque, T, is proportional to the speed, ω.
A is the proportional constant which they calculate as 0.382 Ns.
This is similar to a y=mx relation and they are calculating m from one point on the graph:
m = y/x or in this case A=T/ω
All that needs to be done is to convert the 400 rev/min to rad/s.