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 is

• A. $10\text{J}$
• B. $2\text{J}$
• C. $9.5\text{J}$
• D. $0.5\text{J}$

Answer 1

The correct option is C $9.5\text{J}$

Given $m=4\text{kg}$
Potential energy, $U\left(x\right)=\frac{x^3}{3}-\frac{5x^2}{2}+6x+3$
Given: Total mechanical energy $E=17\text{J}$
We know that,
$E=KE+PE=KE+U$

$\therefore$ For maximum $KE,U$ should be minimum.
i.e $\frac{dU}{dx}=x^2-5x+6=0$
$\Rightarrow (x-3)(x-2)=0$
$\therefore x=3,2$
$\frac{d^{2}U}{d{x}^2}=2x-5$
At $x=3,\frac{d^{2}U}{d{x}^2}=1>0$ (minimum)

$\therefore$ Minimum potential energy at $x=3$
$U(x=3)=\frac{3^3}{3}-\frac{5\times 9}{2}+18+3$
$U_{\text{min}}=7.5\text{J}$
$\therefore (K.E{)}_{\text{max}}=E-U_{\text{min}}$
$=17-7.5=9.5\text{J}$