Question

A particle moves along the x axis from x = 12.8 m to x = 23.7 m under the influence of a force$\mathrm{F}=\frac{375}{\left({\mathrm{x}}^3+3.75\mathrm{x}\right)}$ where F is in newtons and x is in meters. Using numerical integration, determine the work done by this force on the particle during this displacement. Your result should be accurate to within 2%.

Short answer

The work done by force on particle is 0.799 J

Step-by-step solution

Identification of given data

The force on particle is$\mathrm{F}=\frac{375}{\left({\mathrm{x}}^3+3.75\mathrm{x}\right)}$

The initial position of particle is x=12.81 m

The final position of particle is x =23.7 m

Determination of work done by force by numerical technique

The work done by force is given as:

$\mathbf{W}\mathbf{=}\underset{\mathbf{12}\mathbf{.}\mathbf{8}\mathbf{m}}{\overset{\mathbf{23}\mathbf{.}\mathbf{6}\mathbf{m}}{\int }}\mathbf{F}·\mathbf{dx}$

Here, dx is the small change in position of particle and its value is.

Substitute all the values in the above equation.

$$\begin{gathered} W=\left(\frac{375}{{\left(12.8\text{m}\right)}^3+3.75\left(12.8\text{m}\right)}\right)\left(0.100\text{m}\right)+\left(\frac{375}{{\left(12.9\text{m}\right)}^3+3.75\left(12.9\text{m}\right)}\right)\left(0.100\text{m}\right)+....... \\ +\left(\frac{375}{{\left(23.6\text{m}\right)}^3+3.75\left(23.6\text{m}\right)}\right)\left(0.100\text{m}\right) \\ W=0.806\text{J} \end{gathered}$$

The work done by force is given as:

$\mathbf{W}\mathbf{=}\underset{\mathbf{12}\mathbf{.}\mathbf{8}\mathbf{m}}{\overset{\mathbf{23}\mathbf{.}\mathbf{6}\mathbf{m}}{\int }}\mathbf{F}·\mathbf{dx}$

Substitute all the values in the above equation.

$$\begin{gathered} W=\left(\frac{375}{{\left(12.8\text{m}\right)}^3+3.75\left(12.8\text{m}\right)}\right)\left(0.100\text{m}\right)+\left(\frac{375}{{\left(12.9\text{m}\right)}^3+3.75\left(12.9\text{m}\right)}\right)\left(0.100\text{m}\right)+....... \\ +\left(\frac{375}{{\left(23.7\text{m}\right)}^3+3.75\left(23.7\text{m}\right)}\right)\left(0.100\text{m}\right) \\ W=0.791\text{J} \end{gathered}$$

The actual work will lie between these values so the actual work done by force is given as:

$$\begin{gathered} W_a=\frac{0.806\text{J}+0.791\text{J}}{2} \\ {W}_a=0.7985\text{J} \\ {W}_a=0.799\text{J} \end{gathered}$$

Therefore, the work done by force on particle is 0.799J.