i need a solution please

Question

A 1200 kg package is being moved on the floor of a plant at a speed of 10 m/s under

the action of a horizontal force F (figure 2). Suddenly an additional force of 1000

N is applied to the package to retard its motion. Determine the time required

to bring the package's speed to 5 m/s. The coefficient of kinetic friction

between the package and the floor is 0.35.

Answer 1

The package is initially in equilibrium (constant speed) We know the Normal Force = 1200 x 9.8 = 11760 N

The pushing force must be the same as the friction force = 0.35 x 11760 = 4116 N.

(Note we actually don't nee the above but I thought I would include it)

If we now have a retarding force of 1000 N, this will be the new net force = 1000 N

From this the acceleration will be -1000/1200 = -0.833 m/s2

The original velocity is 10 m/s and the final is 5

We can now write 5 = 10 -(0.833)(t)

So t = 6 s