If a box is pushed across a floor with a force of 130n. the frictional force acting between the box and the floor is 30n, over a time period of 5 sec,
determine :
1) what is the net force acting on it?
2) acceleration acting upon the box if it starts from rest and attains a certain kinetic energy after being pushed 25m?
3) also find the mass of the box.
4) what will be the kinetic energy and final velocity?

The force acting in the front direction is the 130N.
The frictional force is acting backwards          30N.

1) The net force is 130N - 30N    =  100N

2)  s  = ut + (1/2)at^2             u = 0,  Start from rest,  s = 25m t =5.

25 = 0*5  +  (1/2)* a * 5^2.

25 = 0  +  25/2  * a.

25  =    (25/2)a.      Divide 25 from both sides.

1 =  (1/2)* a.          Cross multiply.

2 = a.

a = 2 m/s^2.

3) Mass of the box
Net Force,  F = ma
                   100 =  m*2.        Divide both sides by 2.
                    
                     100/2  =  m
                       50       =  m.
                        m  = 50 kg.

4)  Final velocity ,   v = u + at.
                                   v  =  0  + 2*5 = 10 m/s.
                                  
   Kinetic Energy,  K  =  (1/2) * mv^2.
                                    =    1/2  * 50 * 10 * 10.
                                    =      2500 J.

Question was not very clear if there are specific answer choices tell me and i'll try to you more. hope this though2 laps is 1 min. we know that for each min she does two times the amount of laps. for 12 laps we can do 12/2 which does give us 12 laps in 6 mins not 16 mins.