# Question #ffb6c

May 7, 2017

$F = - \frac{\mathrm{dU}}{\mathrm{dx}} \implies U = - \int 6 x - 12 \mathrm{dx} = - 3 {x}^{2} + 12 x + C$

$U \left(0\right) = 20 \implies C = 20$

$\implies U \left(x\right) = 20 + 12 x - 3 {x}^{2}$

So:

$U \left(3\right) = 29 J$

May 7, 2017

$29 J$

#### Explanation:

If a force which acts on an object is a function of position only, it is called a conservative force.
It can be represented by a potential energy function which for a one-dimensional case satisfies the following condition

$- \frac{\mathrm{dU} \left(x\right)}{\mathrm{dx}} = F \left(x\right)$

or in integral form as

$U \left(x\right) = - {\int}_{{x}_{0}}^{x} F \left(x\right) \mathrm{dx}$
where ${x}_{0}$ is the reference point

Inserting given function we get
$U \left(x\right) = - \int \left(6 x - 12\right) \mathrm{dx}$
$\implies U \left(x\right) = - 6 {x}^{2} / 2 + 12 x + C$ .......(1)
where $C$ is constant of integration.

To evaluate $C$ we find the value of function $U \left(x\right)$ at $x = 0$. We have
$- 3 \times {0}^{2} + 12 \times 0 + C = 20$
$\implies C = 20$
From (1) we get
$U \left(x\right) = - 3 {x}^{2} + 12 x + 20$

$\therefore U \left(3\right) = - 3 \times {3}^{2} + 12 \times 3 + 20 = 29 J$