In a tractor-pull competition, a tractor applies a force of 1.3 kN to the sled, which

I assume the tractor pulls completely horizontal

Draw a FBD of the sled.  Include forces from weight (down), from normal force (up), from tractor (right), and from friction (left).

Weight and normal force are equal and opposite, since the sled doesn't rise.  Friction is the coefficient times the normal force.

Standard value of g = 9.8 N/kg

We can conclude the following formulas for each force:
T = given tractor force, convert to regular newtons
W = m*g
N = m*g

Resulting:
Ff = mu*m*g

The net force is Fnet = T - mu*m*g

Fnet = -84940 Newtons

Use Newton's 2nd law:
Fnet = m*a
a = Fnet/m

Resulting acceleration:
a = -7.72 m/sec^2

Negative sign indicates that the sled is SLOWING DOWN.