DC Motor: Difference between E and k?
There are two key constants with respect to electrical machines

1) \$K_e\$ The BackEMF constant. Units in Volts/rpm (volts/rads ...) -> what voltage will be generated at the terminals if the rotor is spinning.

2) \$K_t\$ The Torque constant. Units in Nm/A - What torque will be generated if current is injected into the stator 

(there is a 3rd... \$K_v\$, velocity constant, the reciprocal of \$K_e\$, rpm/Volts) - what speed the rotor could spin at for a given applied voltage

In theory \$K_t\$  = \$K_e\$. In practice, \$K_t\$ &lt; \$K_e\$ as \$K_t\$  is taken at rated current &amp; corresponds to Iron saturation.

So the question: 


  A DC machine has a back-emf constant of 0.24 V/(rad/s) and a
  resistance of 0.5 ohms. It is operating from a 12V DC power supply. At
  its rated speed of 400 rpm, calculate its rated back-emf (that is, the
  back-emf at rated speed).


We thus know \$K_e\$ = 0.24V/rad/s 

We also know the rpm = 400rpm

convert rpm to rad/s: \$ \omega = \frac{rpm * 2 * \pi}{60}  = 41.8879rad/s \$

There is one more step to get the rated terminal voltage, but I will leave that