The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force

which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,

so

The frictional force can be rewritten as

where

,

. Re-arranging, we can solve this equation to find

, the coefficient of dynamic friction:
I think you want to determine the exit speed?
You have to determine how much velocity was decreased by calculating it from the kinetic energy.
KE = (1/2)mv²
1.4 x 10^5 = (1/2)*(1100)v²
v² = 254.55
v =15.95 m/s
So the velocity reduces by 15.95 m/s. Subtracting this to the initial velocity: 22 - 15.95 = 6.05 m/s.
So, the final speed was 6.05 m/s.
I hope I was able to help :)
Explanation:
50 m/s may squared siya sa dun sa may s
<span>D) The spring has potential energy and the hands have kinetic energy. </span>