Question

The potential energy of a $4\text{kg}$ particle free to move along the $x-$ axis is given by $U\left(x\right)=\frac{x^3}{3}-\frac{5x^2}{2}+6x+3$
Total mechanical energy of the particle is $17\text{J}$. Then the maximum kinetic energy in joules is

Answer 1

$U\left(x\right)=\frac{x^3}{3}-\frac{5x^2}{2}+6x+3$
$\therefore F=-\frac{dU}{dx}=-x^2+5x-6=0$
$\Rightarrow (x-2)(x-3)=0\text{i.e.,}x=2\text{or}3$
Now, $\frac{d^{2}U}{d{x}^2}=2x-5$
At $x=2,\frac{d^{2}U}{d{x}^2}<0\text{i.e.,}U\left(x\right)\to {\text{max}}^m$
At $x=3,\frac{d^{2}U}{d{x}^2}>0\text{i.e.,}U\left(x\right)\to {\text{min}}^m$
$\therefore U^{\text{min}}=\frac{27}{3}-\frac{5\times 9}{2}+6\times 3+3=7.5\text{J}$
$(\text{K.E}{)}^{\text{max}}=17-7.5=9.5\text{J}$