Is my answer to this net force problem right

The question: "A crate is being pulled by a rope which is horizontal to the floor. The crate’s mass is 45.0 kg and it is currently accelerating at +3.50 m/s². The coefficient between the floor and the crate is 0.300. What is the amount of the net force acting on the crate?"

First I found the friction: 0.300 * 441 N = 132 N
then I found the force (applied?): 45 kg * 3.50 m/s^2 = 157 N
then I subtracted them to find Fnet: 157 N - 132 N = 25 N

If anyone has time, could you please check this second part of the same problem?
"What is the tension on the rope? (Hint: What is the applied force, and what is pulling the crate?)"
Would the answer to this be 25 N as well? Somehow I don't think that's right though.
Can someone explain how to do these problems if I got them wrong?
Thanks

The applied force isn't F = ma. The sum of all forces acting in the direction of acceleration is Fnet = ma.

The sum of the forces are:
Fapp+ ( - Ffr) = ma,
Where Fapp is the applied force. Ffr is the force due to friction. And these are equal to mass times acceleration.

The reason why we chose the friction force to be negative is because it is opposite to the direction of acceleration.



Fapp = ma + Ffr
Fapp = 45*3.5 + 132 would give the applied force.

The amount of net force will be simply m*a.

The tension is the same as the force applied.

Gurrl, u a Harvard student!! U go boo.