Answer
The braking force on each rear wheel is $~~960~N$
Work Step by Step
In part (b), we found that the normal force on each rear wheel is $~~F_N = 2400~N$
We can find the braking force on each rear wheel:
$F = F_N~\mu_k$
$F = (2400~N)(0.40)$
$F = 960~N$
The braking force on each rear wheel is $~~960~N$