Here we have to understand the scenario. When the computer drops on to the belt it has no velocity. But gradually it tries to take the velocity of the belt. The friction force acting on the computer will disturb it from getting the velocity of the belt. At this time the computer will drag along the belt. Once it gets the belt velocity it travels freely along the belt under kinetic friction.
Consider the velocity of the computer relative to the belt. Once it drop on the belt it has 3.1m/s velocity. When it smoothly running on belt it was 0m/s.
Using `F = ma` to the belt running direction.
At the dragging phase the only force acting on the computer is friction fore.
Friction force `= 0.5*10*9.81 = 49.1N`
`-49.1 = 10*a`
`a = -0.491`
Using `V^2 = U^2+2aS`
`0 = 3.1^2-2*0.491*S`
`S = 0.98m`
So the computer drags for 0.98m before runs smoothly.