Answer:
d) 1.44m
Explanation:
According to Newton's second law:
[tex]\sum F=m.a\\[/tex]
analyzing the horizontal components of the force:
[tex]\sum F_x=mg*sin(\theta)-\µ*m.g*cos(\theta)\\\sum F_x=m*9.8*(sin(16^o)-0.420*cos(16^o))\\\sum F_x=-1.25m[/tex]
applying the second law
[tex]-1.25m=m.a\\a=-1.25m/s^2[/tex]
given the acceleration we can calculate the distances traveled before stopping:
[tex](v_f)^2=(v_o)^2+2.a.\Delta x\\0=(1.90m/s)^2-2(1.25)*\Delta x\\\Delta x=1.44m[/tex]