The net work done on the cart can be given as,
[tex]W=Fd+W_f[/tex]The change in kinetic energy of the cart is,
[tex]\Delta K=\frac{1}{2}m(v^2-u^2_{})[/tex]According to work energy theorem,
[tex]W=\Delta K[/tex]Substitute the known expression,
[tex]\begin{gathered} Fd+W_f=\frac{1}{2}m(v^2-u^2) \\ v^2-u^2=\frac{2(Fd+W_f)}{m} \\ v^2=\frac{2(Fd+W_f)}{m}+u^2 \end{gathered}[/tex]Substitute the known values,
[tex]\begin{gathered} v^2=\frac{2((4.29\text{ N)(34.0 cm)(}\frac{1\text{ m}}{100\text{ cm}})\text{-0.237 J)}}{(530\text{ g)(}\frac{1\text{ kg}}{1000\text{ g}})}+(1.30m/s)^2 \\ =4.610m^2s^{-2}+1.69m^2s^{-2} \\ v=\sqrt[]{6.300m^2s^{-2}} \\ \approx2.51\text{ m/s} \end{gathered}[/tex]Thus, the final speed of cart is 2.51 m/s.
