Figure shows the forces acting on 2 kg block.
In vertical direction, we have, N + F sin37 = 2g .........(1)
where N is normal reaction force, we get N = 2g- Fsin37 = 2×9.8 - 20×.6 = 7.6 Newton
hence in horizontal direction, we have, F×cos37 - μ×N = m×a .........(2)
where μ is the coefficent of friction and a is acceleration
we substitute values in eqn.(2) and get a = 7.05 m/s2
Starting from rest, if the block moves 8 m with acceleration a, then speed v of block after travelling 8 m is given by ,
after moving 8m distance, if the horizontal component reversed, the applied force gives retardation to the movement of block.
when the horizontal component is reversed but still the block is moving implies that
frictional force and applied force both are in same direction. hence retardation ar is given by
F×cos37 + μ×N = m×ar .....................(3)
Hence ar = 8.95 m/s2
Distance travelled before stopping :- S = u2 / (2ar ) = (10.62×10.62)/(2×8.95) = 6.3 m