No, no problem! I would be happy if I could help. I should also admit that our other friend Onlineslayer was also helpful as my reply was something of a quick response.
Part e was a bit more complicated than initially I thought it was. The assumption is that even if the braking force is not applied, both reindeer and the sleight will decelerate at the same rate of 0.25m/s^2 (This was the assumption you made when calculating the distance travelled by the reindeer and sleight as you essentially could consider the reindeer and sleigh as one body even though they were connected by a rigid harness). This means you have already taken account of the resistance to motion of the reindeer when calculating the deceleration (or the distance travelled) so you will have to bear this in mind for the next part.
Although the deceleration of the reindeer results from the tension in the harness as well as the resistance of the reindeer itself, you should not take into account the resistance of the reindeer when considering the tension in the harness calculated from the equation
F=ma, where
a will be the deceleration of the reindeer, as you have already considered this force when calculating the deceleration. You could also calculate the tension in the harness by considering the
work done by the tension and the kinetic energy of the reindeer, that is,
21mv2=f×d. Again, note that the force in this case is only the tension in the harness without the resistance force because of the consideration of forces when calculating the distance trevelled.
I hope this is correct.
Edit: I've had a couple of glasses of wine (regardless of the fact that it's actually too late) so the thought that I might be wrong is crossing my mind.
Second edit (
@coconut64): indeed I was wrong in saying that the force in each case is the tension in the harness only, it is the tension as well as the resistance to the motion of the reindeer. Please do ask questions if you still have doubt about it.