Question

A mass m = 16 kg is pulled along a horizontal floor with NO friction for a distance d =8.4 m. Then the mass is pulled up an incline that makes an angle ? = 25

Answer 1

Friction = mu*N
In this case N = m*g*cos35
The second neglected to consider gravitational potential energy gain due to going a distance up the incline. And also, at least in the write-up, left out the mass when calculating friction.

I'll follow the first approach (with correction) and use my nomenclature while working the askmehelpdesk problem. Along the incline, the forces that apply are friction, weight and the pulling force. Since we know that it comes to a stop, I'll declare that down the incline is the positive direction.

Fnet = friction + downslope component of weight - tension
Fnet = Ff + m*g*sin35 - T = mu*m*g*cos35 + m*g*sin35 - 55 N
Fnet = 0.34*12 kg*9.8 m/s^2*cos35 + 12 kg*9.8 m/s^2*sin35 - 55 N
Fnet = 32.8 N + 67.5 N - 55 N = 45.3 N
Fnet = m*a = 45.3 N
a = F/m = 45.3 N/12 kg = 3.77 m/s^2

The steps taken by the first response in askmehelpdesk after obtaining acceleration are correct. I'll work that out since I need it for the next part.
Vf^2 = Vi^2 + 2*a*d
-2*a*d = Vi^2
-2*3.77 m/s^2*d = (6.249 m/s)^2 = 39.1 m^2/s^2
d = -5.18 m

Second part. Original energy is all kinetic
KEi = (1/2)m*v^2 = (1/2)*12 kg*(6.249 m/s)^2 = 234 J
Ending energy is all gravitational potential energy
GPE = m*g*h = 12*g*5.18 m*sin35 = 35.7 J
The mass lost energy, so negative work was done.

I'm quite sure of my method but you should check my arithmetic and substitutions