A 39.1 kg woman wearing ice skates is standing on a skating rink next to a vertical wall. She pushes off against the wall and begins to slide along the perfectly horizontal surface of the rink. The coefficient of kinetic friction between the blades of her ice skates and the surface of the rink is 0.04. If she slides a distance of 14.8 m before stopping and her push against the wall lasted 0.2 s, what was the average force (in N) that she exerted on the wall?

666.67N

Explanation:

from work energy principle

work done by friction = change in KE

-uk*m*g*x = (1/2)*m*(vf^2-vi^2)

-uk*g*x = (1/2)*(vf^2-vi^2)

as the woman is stopped the final velocity vf = 0

-0.04*9.8*14.8 = (1/2)*(0-vi^2)

initial speed vi = 3.41 m/s

after pushing against the wall

momentum gained by woman = m*vi

force applied F = m*vi/t

F = 39.1*3.41/0.2

F = 666.66 N <<

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.