Beginning at time t = 0, a student exerts a horizontal force on a box of mass 30 kg, causing it to move at 1.2 m/s toward an elevator door located 16 m away, as shown above. the coefficient of kinetic friction μk between the box and the floor is 0.20. calculate the magnitude of the horizontal force the student must exert on the box in order to keep it moving at 1.2 m/s.

Since the initial and the final velocity are just equal, it is implied that the acceleration is zero. This means that the net force acting on the body is also zero. The horizontal force should be equal to the force of friction. The force of friction is equal to the product of the coefficient and the normal force. 
                                 Ff = (coefficient of friction) x Fn
The normal force is equal to the object's weight if the surface is horizontal.
                               Ff = (0.20) x (30 kg) x (9.8 m/s²)  = 58.8 N
Thus, the horizontal force exerted must be 58.8 N. 

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