Eve pushes a 1.2kg box, initially at rest, across a flat surface with a constant

Question

Eve pushes a 1.2kg box, initially at rest, across a flat surface with a constant, horizontal force of 8.0N. After traveling an unknown distance, she stops pushing the box. The box continues along in a straight line until it comes to a stop 150m away from where she began pushing it. The coefficient of kinetic friction between the box and the floor is 0.25.
What is the total distance that Eve pushes the box for?
I have found the acceleration of the box while pushing it and while it is coming to a stop.
,push=4.22 m/s2      

,friction=-2.45m/s2

I do not know how to set up the equation so that I can find the distance she pushes the box for. Please help.

Explanation / Answer

kinetic energy of the box just before stopping the push = work done by friction to stop it

kinetic energy of box just before stopping the push  = work done by net force = maS =m x 4.22 x S

work done by friction to stop it = force due friction x 150 = m x 2.45 x 150

equating above two, 4.22 x S = 2.45 x 150

S =  87.08 m

N.B you need not find the acceleration of individual motions, just equate the work done by net force as 8-mg in first equation, and friction = mg in second equation, u get the same result.

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