# Question 69321

##### 1 Answer
Nov 29, 2017

$W = - \left[\frac{a}{2} \left({x}_{f}^{2} - {x}_{i}^{2}\right) + \frac{b}{3} \left({x}_{f}^{3} - {x}_{i}^{3}\right)\right] = - 51.08$ $m J$
U(x) = a/2x^2 + b/3x^3;
${U}_{f} = \frac{a}{2} {x}_{f}^{2} + \frac{b}{3} {x}_{f}^{3} = 0.05737$ $J = 57.37$ $m J$

#### Explanation:

Work Done Vs Potential Energy: For conservative forces, the work done by the force can be written as the negative of change in potential energy.

$W = - \setminus \Delta U = - \left({U}_{f} - {U}_{i}\right)$ ...... (1)

F(x) = -ax -bx^2; \qquad a = 5.00  Nm^{-1}; \qquad b = 1.00 Nm^{-2};#

${x}_{i} = 5.0$ $c m = 5.0 \setminus \times {10}^{- 2}$ $m$;
${x}_{f} = {x}_{i} + \setminus \Delta x = 5.0$ $c m + 10.0$ $c m = 15.0$ $c m = 15 \setminus \times {10}^{- 2}$ $m$.

Work Done:

$W = \setminus {\int}_{{x}_{i}}^{{x}_{f}} F \left(x\right) \mathrm{dx} = \setminus {\int}_{{x}_{i}}^{{x}_{f}} - a x - b {x}^{2} \mathrm{dx}$
$W = - {\left[\frac{a}{2} {x}^{2} + \frac{b}{3} {x}^{3}\right]}_{{x}_{i}}^{{x}_{f}}$
$W = - \left[\frac{a}{2} \left({x}_{f}^{2} - {x}_{i}^{2}\right) + \frac{b}{3} \left({x}_{f}^{3} - {x}_{i}^{3}\right)\right]$ ...... (2)

Substituting for $a$, $b$, ${x}_{i}$ and ${x}_{f}$,

$W = - \left(0.05000 + 0.00108\right) = - 0.05108$ $J = - 51.08$ $m J$

Potential Energy: We are not asked to calculate the change in potential energy. Rather we have been asked to calculate the final potential energy.

Comparing (1) with (2), we conclude that the initial and final potential energies are,

${U}_{i} = \frac{a}{2} {x}_{i}^{2} + \frac{b}{3} {x}_{i}^{3} = 0.00629$ $J = 6.29$ $m J .$
${U}_{f} = \frac{a}{2} {x}_{f}^{2} + \frac{b}{3} {x}_{f}^{3} = 0.05737$ $J = 57.37$ $m J$

For a displacement $x$, in general, $U \left(x\right) = \frac{a}{2} {x}^{2} + \frac{b}{3} {x}^{3}$