Two 50kg sleds connected by a rope are pulled horizontally at constant velocity by four dogs across ice with negligible friction so that the tension in the rope is 200N. What is the work performed by the dogs over 100m?

The answer says:

W = (Fcosθ)d

W = Fd since the force and displacement are in the same direction

The displacement is given as 100m but to solve for the work performed by the dogs their pulling force must be determined.

The acceleration of the box on the left can be calculated from Newton's 2nd law applied in the horizontal direction:

SF = ma

200N = (50kg)a

a = 4m/s2

As the two blocks are tethered to each other and move together this will also be the acceleration of the box on the right, and we can use Newton's 2nd law applied in the horizontal direction to the box on the right to calculate the dogs' pulling force.

SF = ma

Fdogs - 200N = (50kg)(4m/s2)

Fdogs = 400N

W = Fd

W = (400N)(100m)

W = 40,000J = 40kJ

But in the question, it says that the two sleds travel at

__constant velocity,__so wouldn't acceleration be zero? Can someone help explain this to me? Thanks so much in advance.