The box has weight 50.0 N (a downward force), from which we can determine its mass [tex]m[/tex]:
[tex]-50.0\,\mathrm N=m(-g)=m\left(-9.80\,\dfrac{\mathrm m}{\mathrm s^2}\right)\implies m=5.10\,\mathrm{kg}[/tex]
The box's acceleration is taken to be uniform, which means its acceleration due to the frictional force (which acts in the leftward direction) at any time during the [tex]\Delta t=t_2-t_1=2.25\,\mathrm s[/tex] interval is
[tex]\vec a=\dfrac{0-1.75\,\frac{\mathrm m}{\mathrm s}}{2.25\,\mathrm s}=-0.778\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]
Then the friction force has magnitude [tex]F[/tex] (where the vector is acting in the leftward direction) satisfies
[tex]\vec F=m\vec a\implies-F=(5.10\,\mathrm{kg})\left(-0.778\,\dfrac{\mathrm m}{\mathrm s^2}\right)\implies F=3.97\,\mathrm N[/tex]
and the closest answer would be A.
