coefficient of friction formulas

A 60kg boy is on a 15kg sled. He is pushed along a level path of snow where uk(coefficient of kinetic friction)=0.05.
What total force will accelerate it at 1.3 m/s/s?
Using an answer to a previous problem my teacher gave me as a paradigm, I think I know how to get the answer:
Ff=(.05)(75*9.81)=36.78
Fapplied=36.78+(75*1.3)
Fapplied=134N
Is this correct. If not, please explain why.
But what is wrong with this formula:
force of friction= m(-a).
Ff=(.05)(75)(-1.3).
When do I and when don't I use this above formula.
Here is an example of when my teacher used that formula:
A 10 kg mass is moving at 10m/s on a level surface. The kinetic coefficient of friction is .13. My teacher used that formula to find the acceleration in this problem.
Why can't i use this formula in the previous problem?
Thank you guys for your help.!