Question

A net force along the x-axis that has x-component ${\mathbf{F}}_{\mathbf{x}}\mathbf{=}\mathbf{[}\mathbf{-}\mathbf{20}\mathbf{\hspace{0.33em}}\mathbf{N}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{300}\mathbf{\hspace{0.33em}}\mathbf{N}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}\mathbf{\left)}{\mathbf{x}}^{\mathbf{2}}\mathbf{\right]}$is applied to an 5.0 kgobject that is initially at the origin and moving in the -x-direction with a speed of 6.0m/s. What is the speed of the object when it reaches the point x=5.00 m.

Short answer

At a distance of 5m from the origin, the velocity of the object is 4.12 m/s

Step-by-step solution

Identification of given data

The given data can be listed below,

• The net force is,${\mathrm{F}}_{\mathrm{x}}=-12.0\hspace{0.33em}\mathrm{N}+(0.300\hspace{0.33em}\mathrm{N}/{\mathrm{m}}^2){\mathrm{x}}^2$
• The mass of the object is $\mathrm{m}=5.0\hspace{0.33em}\mathrm{kg}$,
• The speed of the object is, v=6 m/s

Concept/Significance of velocity

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.

Determination of the speed of the object when it reaches the point.

The work done by the force $F_y$in moving from $x=0\hspace{0.33em}to\hspace{0.33em}x=x_{stop}$is given by,

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

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,

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

The kinetic energy of the object is given by,

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

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

Substitute the given values in the above,

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

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