Given data:
Constant horizontal force,
[tex]F=4\times10^3\text{ N}[/tex]Mass of the object,
[tex]m=10\times10^3\text{ kg}[/tex]Initial velocity of the object,
[tex]u=8.0\text{ m/s}[/tex]Distance moved;
[tex]s=150\text{ m}[/tex]Free body diagram when surface is smooth;
The force is given as,
[tex]F=ma[/tex]Here, a is the acceleration of the object.
The expression for the acceleration of the object is given as,
[tex]a=\frac{F}{m}[/tex]Substituting all known values,
[tex]\begin{gathered} a=\frac{4\times10^3\text{ N}}{10\times10^3\text{ kg}} \\ =0.4\text{ m/s}^2 \end{gathered}[/tex]From third equation of motion the final velocity is given as,
[tex]v=\sqrt[]{u^2+2as}[/tex]Substituting all known values,
[tex]\begin{gathered} v=\sqrt[]{(8.0\text{ m/s})^2+2\times(0.4\text{ m/s}^2)\times(150\text{ m})} \\ \approx13.56\text{ m/s} \end{gathered}[/tex]Therefore, the final velocity of object when it moved on smooth surface is 13.56 m/s.
(ii)
When the body is moved on the rough surface. The amount of frictional force acting on the object is,
[tex]f=200\text{ N}[/tex]Free body diagram when the surface is rough:
The net force acting on the body is given as,
[tex]F_{net}=F-f_{}^{}[/tex]Substituting all known values,
[tex]\begin{gathered} F_{net}=(4\times10^3\text{ N})-(200\text{ N}) \\ =3800\text{ N} \end{gathered}[/tex]The net force acting on the body is given as,
[tex]F_{net}=ma^{\prime}[/tex]Here, a' is the new acceleration of the object.
The expression of the new acceleration is given as,
[tex]a^{\prime}=\frac{F_{net}}{m}[/tex]Substituting all known values,
[tex]\begin{gathered} a^{\prime}=\frac{3800\text{ N}}{10\times10^3\text{ kg}} \\ =0.38\text{ m/s}^2 \end{gathered}[/tex]From third equation of motion new velocity of the object is given as,
[tex]v^{\prime}=\sqrt[]{u^2+2a^{\prime}s}[/tex]Substituting all known values,
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