Answer
$3\sqrt{2}\approx 4.24 \ m/s$
Work Step by Step
The work done can be calculated as:
$ Work \ done =\int_{0}^{3} (3x-x^2) \ dx \\=[\dfrac{3x^2}{2}-\dfrac{x^3}{3}]_0^3\\=\dfrac{3(3^2) }{2}-\dfrac{(3^3) }{3} \\=4.5 \ J$
The kinetic energy equals the work done. So,
$4.5 =\dfrac{1}{2} m[v(t_2)]^2-(v(t_1))^2] $
Since, $v(t_1)=0$
Therefore,
$v(t_2) =\sqrt {18} =3\sqrt{2}=4.24 \ m/s$
