(a) Derive an expression for the potential energy of an ob subject to a force \(F_{x}=a x-b x^{3},\) where \(a=5 \mathrm{N} / \mathrm{m}\) \(b=2 \mathrm{N} / \mathrm{m}^{3},\) taking \(U=0\) at \(x=0 .\) (b) Graph the poten energy curve for \(x>0\) and use it to find the turning points an object whose total energy is \(-1 \mathrm{J}\)

Understanding the relationship between force and potential energy is crucial for students studying physics. In essence, potential energy is the stored energy in an object due to its position or configuration. When we examine the relationship with force, we find a fundamental concept: potential energy change in a system is equal to the negative work done by the force as the object moves from one point to another.

Let's take a force represented by the equation \( F_x = ax - bx^3 \), where \( a \) and \( b \) are constants. The work done moving from one point to another involves integrating this force over the distance moved. Since the potential energy (U) is defined as the negative of this work, we have \( U(x) = - \int F(x) \, dx \). This negative sign reflects the idea that if a force does positive work on an object, it loses potential energy, and vice versa.

To determine an object's potential energy mathematically, we need to integrate the force over displacement. For the given force \( F_x = ax - bx^3 \), integration is the method of finding the antiderivative or the area under the force-displacement graph.

The equation for potential energy after integration is obtained by evaluating \( U(x) = - \int (ax - bx^3) \, dx \), which, with our specific values for \( a \) and \( b \), simplifies to \( U(x) = -\frac{5x^2}{2} + \frac{x^4}{2} \) when taking into consideration that the potential energy is zero when \( x = 0 \). This process translates a force into a potential energy function, which can be visualized in a graph and analyzed for physical predictions.

Graphing the potential energy curve of an object as a function of its position provides invaluable insights into the object's dynamics without actually watching it move. By plotting \( U(x) \) which we computed as \( -\frac{5x^2}{2} + \frac{x^4}{2} \), we can visualize how the potential energy changes with position.

The shape of the potential energy curve can tell us about the stability of an object's position. For instance, a ball resting at the bottom of a bowl has minimum potential energy and is in a stable equilibrium. If the curve has dips (minima), they represent stable positions. Peaks (maxima), conversely, indicate unstable positions. By carefully analyzing the curve, we can gain a deeper understanding of the physical phenomena at play.

Turning points on a potential energy curve are significant because they indicate positions where the object can potentially change direction. These points occur where the derivative of the potential energy with respect to position, \( \frac{dU}{dx} \), is zero. This means that an object at a turning point has no net force acting on it and is momentarily at rest from a mechanical perspective.

For the problem at hand, we would set the derivative of \( U(x) \) to zero and solve for \( x \). Turning points also correspond to the energy levels where an object can have potential energy equal to its total mechanical energy. In our exercise, we have a total energy of \(-1 \mathrm{J}\), which means by solving the equation \( -\frac{5x^2}{2} + \frac{x^4}{2} = -1 \), we can determine the turning points, offering valuable insights into the motion of the object at particular energy states.