The tractor moved evenly on a horizontal road, developing a traction force of 4 kN. When a sled weighing

The tractor moved evenly on a horizontal road, developing a traction force of 4 kN. When a sled weighing 800 kg was attached to it, the traction force with uniform movement had to be increased to 6 kN. Find the coefficient of friction between the sled runners and the road.

given
pulling force F pulling = 4 kN
sleigh weight m = 800 kg
increased traction force F traction = 6 kN
find the coefficient of friction
decision
there are four forces at work:
gravity m * g
support reaction force N
traction force F
friction force F tr
using Newton’s first law N + m * g + Fthrust + Ffr = 0
on the 0x axis: F thrust = F tr
on the 0y axis: N = m * g;
mnu – coefficient of friction
F thrust = F tr = mnyu * N = mnyu * m * g
friction coefficient = 6 k N / (800 kg * 10 N / kg) = 6 * 1000 / (800 * 10) = 6/800 * 100 = 6 * 100/800 = 6 * 100 / (8 * 100) = 6 / 8 = 3/4
the answer is 6 3/4