Q68P - University Physics with Modern Physics

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 kg object 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.

Step-by-Step Solution

Answer

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

Step 1: 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  

Step 2: 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.

Step 3: 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, m is 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