91Ó°ÊÓ

The given data can be listed below,

• The net force is,${\mathrm{F}}_{\mathrm{x}}=-12.0 \mathrm{N}+(0.300 \mathrm{N}/{\mathrm{m}}^2){\mathrm{x}}^2$
• The mass of the object is $\mathrm{m}=5.0 \mathrm{kg}$,
• The speed of the object is, v=6 m/s

When a moving system is seen by a mostly stationary observer, velocity is a vector that reflects the displacement between two places in space with respect to time.

The work done by the force $F_y$in moving from $x=0 to x=x_{stop}$is given by,

$\mathrm{W}(0,{\mathrm{x}}_{\mathrm{stop}})={∫}_0^{{\mathrm{x}}_{\mathrm{stop}}}\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx}$

The work done by the force$F_x$in moving from$\mathrm{x}={\mathrm{x}}_{\mathrm{stop}}\mathrm{to}\mathrm{x}={\mathrm{x}}_1$is given by,

$W(x_{stop},x_1)={∫}_{x_{stop}}^{x_1}F\left(x\right)dx$

The kinetic energy of the object is given by,

$$\begin{gathered} K\left(x_1\right)=W\left(0,x_{stop}\right)+W\left(x_{stop},x_1\right)+\frac{1}{2}m{v}_0^2 \\ K\left(x_1\right)={∫}_0^{x_1}F\left(x\right)dx+\frac{1}{2}m{v}_0^2 \end{gathered}$$

Here,$v_0$is the initial velocity of the object.

Substitute the given values in the above,

$$\begin{gathered} \mathrm{K}\left({\mathrm{x}}_1\right)={∫}_0^{{\mathrm{x}}_1}\left(-12.0\mathrm{N}+\left(0.300\mathrm{N}/{\mathrm{m}}^2\right){\mathrm{x}}^2\right)\mathrm{dx}+\frac{1}{2}{\mathrm{mv}}_0^2 \\ ={\left(\frac{1}{3}\left(0.300\mathrm{N}/{\mathrm{m}}^2\right){\mathrm{x}}^3-\left(12.0\mathrm{N}\right)\mathrm{x}\right)}_0^{5\mathrm{m}}+\frac{1}{2}\left(5\mathrm{kg}\right){\left(6\mathrm{m}/\mathrm{s}\right)}^2 \\ =\left(\frac{1}{3}×\left(0.300\mathrm{N}/{\mathrm{m}}^2\right){\left(5\mathrm{m}\right)}^2-\left(12.0\mathrm{N}\right).\left(5\mathrm{m}\right)+90\mathrm{J}\right) \\ =42.5\mathrm{JA} \end{gathered}$$

The speed of object when it reaches a distance is given by,

$v\left(x_1\right)=\sqrt{\frac{2K\left(x_1\right)}{m}}$

Here,$K\left(x_1\right)$ is the kinetic energy of the object after a distance, mis the mass of the object.

Substitute all the values in the above,

$$\begin{gathered} \mathrm{v}\left({\mathrm{x}}_1\right)=\sqrt{\frac{2\left(42.5\mathrm{J}\right)}{5\mathrm{kg}}} \\ =4.12\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity of the object after reaching a point 5m is 4.12 m/s