A force $\overrightarrow{\mathbf{F}}\mathbf{=}\left(\mathrm{cx}-3.00x^2\right)\hat{\mathbf{i}}$acts on a particle as the particle moves along an x axis, with$\overrightarrow{\mathbf{F}}$in newtons, x in meters, and c a constant. At x=0, the particle’s kinetic energy is 20 .0 J; at x=3.00 m, it is 11.0 J. Find c.

1. The applied force, $\mathrm{F}=\left(\mathrm{cx}-3.00{\mathrm{x}}^2\right)\hat{\mathrm{i}}\mathrm{N}$
2. At x=0, the kinetic energy of particle is, ${\mathrm{K}}_{\mathrm{i}}=20.0\mathrm{J}$
3. At x=3.0, the kinetic energy of particle is, ${\mathrm{K}}_{\mathrm{f}}=11.0\mathrm{J}$

We can use the equation ofthework-energy theorem to find constant . As the energy at both points is given, we can get the value of using the equation for work in the work-energy theorem.

Formula:

The equation of work-energy theorem

$\mathbf{W}\mathbf{=}{\mathbf{K}}_{\mathbf{f}}\mathbf{-}{\mathbf{K}}_{\mathbf{i}}$ …(1)

$\mathbf{W}\mathbf{=}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{i}}}^{{\mathbf{x}}_{\mathbf{f}}}\mathbf{F}\mathbf{d}\mathbf{x}$ ...(2)

The equation of work-energy theorem is

$\mathrm{W}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}$

From the equations 1 and 2.

${\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}={\int }_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{F}d\mathrm{x}$

Substitute all the value in the above equation.

$\begin{array}{rcl}11.0\mathrm{J}-20.0\mathrm{J}& =& {\int }_0^{3.0}\left(\mathrm{cx}-3.00{\mathrm{x}}^2\right)d\mathrm{x}\\ -9.00\mathrm{J}& =& {\left[\frac{{\mathrm{cx}}^2}{2}-3.00\frac{{\mathrm{x}}^3}{3}\right]}_0^{3.0}\\ -9.00\mathrm{J}& =& 4.50\mathrm{c}-27.0\\ \mathrm{c}& =& 4.00\mathrm{N}/\mathrm{m}\end{array}$

Therefore, the value of c is 4.00 N/m.