Question

A single conservative force F = (7.0x - 11) N, where x is in meters, acts on a particle moving along an x axis. The potential energy U associated with this force is assigned a value of 26 J at x = 0. (a) What is the maximum positive potential energy? At what (b) negative value and (c) positive value of x is the potential energy equal to zero?

Answer 1

Answer:

(a) 34.6429J

(b) -1.57 m

(c) 4.71 m

Explanation:

The derivative of the potential energy with respect to the position is equal to the negative of the force, so:

$-\frac{dU}{dx}=F\\ dU=-Fdx$

Then, if we integer both sides, we get:

∫dU = -∫(7x - 11)dx

$U=\frac{-7}{2}x^{2} + 11x + c$

we know that U is equal to 26 J when x is zero, so:

$U=\frac{-7}{2}x^{2} + 11x + c$

$26=\frac{-7}{2}(0)^{2} + 11(0) + c$

$26=c$

Finally, the equation for the potential energy is:

$U(x)=\frac{-7}{2}x^{2} + 11x + 26$

Therefore, the maximum positive potential energy is the energy when x is equal to 11/7. That is because the equation of U is the equation of a parable, and the vertex in a parable is given by:

$x=\frac{-b}{2a} = \frac{-11}{2(-7/2)} =\frac{11}{7}$

Where b is the number beside x and a is the number beside $x^{2}$, Then, the value of maximum U is:

$U(11/7)=\frac{-7}{2}(11/7)^{2} + 11(11/7) + 26$

$U(11/7)=34.6429J$

On the other hand, the negative and positive values of x where the potential energy is equal to zero is calculated as:

$U(x)=\frac{-7}{2}x^{2} + 11x + 26$

$0=\frac{-7}{2}x^{2} + 11x + 26$

if we solve this using the quadratic equation, we get:

$x =\frac{-11+\sqrt{11^{2}-4(-7/2)(26)} }{2(-7/2)}=-1.5747$

$x =\frac{-11-\sqrt{11^{2}-4(-7/2)(26)} }{2(-7/2)}=4.7175$

Finally, the negative and positive values of x where the potential energy is equal to zero are -1.5747 and 4.7175 respectively.