Question

The potential energy of an object is given by U(x)= 8x^2-x^4, where U is in joules and x is
in meters.
a) Determine the force acting on this object.
b) At what positions is this object in equilibrium?
c) Which of these equilibrium positions are stable and which are unstable?

Answer 1

Answer:

he potential energy of an object is given by U(x)= 8x^2-x^4, where U is in joules and x is

in meters.

a) Determine the force acting on this object.

b) At what positions is this object in equilibrium?

c) Which of these equilibrium positions are stable and which are unstable?

Explanation:

Answer 2

(a). The variable force acting on the object is $-16x+4x^3$

(b). The equilibrium position of the object is  x = 0, x = 2 m and x = -2 m. (c). The stable position are at x = 2 and x= -2 . The unstable position is at x = 0

Explanation:

Given that,

Potential energy of an object is $U(x)=8x^2-x^4$

(a). We need to calculate the force acting on this object

Using formula of force

$F=-\dfrac{\partial U(x)}{\partial x}$

$F=-\dfrac{\partial}{\partial x}(8x^2-x^4)$

$F=-16x+4x^3$

(b). We need to calculate the positions of the object in equilibrium

If the object is in equilibrium

Then , F = 0

So,

$-16+4x^2=0$

$4x^2=16$

$x=\pm 2\ m$

The equilibrium position of the object is  x = 0, x = 2 m and x = -2 m.

(c). We find the equilibrium positions

Here, x = 0 is unstable

x = 2 m is stable

x = -2 m is stable.

Hence,

(a) The variable force acting on the object is $-16x+4x^3$

(b) The equilibrium position of the object is  x = 0, x = 2 m and x = -2 m.

(c) The stable position are at x = 2 m and x= -2 m .

The unstable position is at x = 0

Learn more :

Topic : potential energy

https://brainly.in/question/3136277