Problem 60 - Master Resource Book in JEE Main Physics

Question

A particle is placed at the origin and a force \(F=k x\) is acting on it (where \(k\) is a positive constant). If \(U(0)=0\), the graph of \(U(x)\) versus \(x\) will be, figure (where \(U\) is the potential energy function) [UP SEE 2004]

Step-by-Step Solution

Answer

The graph of \( U(x) \) versus \( x \) is a downward-opening parabola.

Step 1: Understanding the Force Function

The force acting on the particle is given by the equation \( F = kx \). This implies that the force is directly proportional to the displacement \( x \), and the constant of proportionality is \( k \).

Step 2: Relate Force to Potential Energy

Force is also related to potential energy \( U \) by the relationship \( F = -\frac{dU}{dx} \). This means that the force can be expressed as the negative derivative of potential energy with respect to position \( x \).

Step 3: Find the Derivative of Potential Energy

Using the relation \( F = -\frac{dU}{dx} = kx \), we can write the equation for the derivative of \( U \) as \( \frac{dU}{dx} = -kx \).

Step 4: Integrate to Find Potential Energy

Integrate \( \frac{dU}{dx} = -kx \) to find \( U(x) \). The integration yields \( U(x) = -\frac{k}{2}x^2 + C \), where \( C \) is a constant of integration.

Step 5: Use Initial Condition to Find Constant

Since \( U(0) = 0 \) is given, substitute \( x = 0 \) into \( U(x) = -\frac{k}{2}x^2 + C \) to find \( C = 0 \). So, the potential energy function simplifies to \( U(x) = -\frac{k}{2}x^2 \).

Step 6: Graphing Potential Energy

The equation \( U(x) = -\frac{k}{2}x^2 \) is that of a downward opening parabola. This indicates that the graph of \( U(x) \) vs. \( x \) will be a parabola that opens downwards with its vertex at the origin.

Key Concepts

Force Function

The concept of a force function is fundamental in physics and helps us understand how forces interact with objects. In our given exercise, the force function is defined as \(F = kx\).

• Here, \(F\) stands for the force exerted on the particle.
• The variable \(x\) represents the displacement from the origin.
• \(k\) is a constant and indicates how strongly the force increases with displacement.

This equation indicates that the force is directly proportional to the displacement, meaning as the particle moves further from the origin, the force increases if \(k\) is positive.
This relationship is typical in situations involving springs and Hooke's Law, where the force is a restoring force trying to bring the system back to equilibrium.

Potential Energy Function

Potential energy is energy stored in an object because of its position or configuration. For the exercise, we need to derive the potential energy function \(U(x)\) from the force function. For this, we utilize the relationship:
\[ F = -\frac{dU}{dx} \]
This tells us that force \(F\) is the negative derivative of \(U\) with respect to \(x\), which helps us find how the potential energy changes as the displacement changes.
When we substitute the given force into this equation, we get:
\[ \frac{dU}{dx} = -kx \]
Integrating this equation provides the expression for potential energy \(U(x)\). The integration gives us:
\[ U(x) = -\frac{k}{2}x^2 + C \]
The constant \(C\) is determined by initial conditions. The exercise specifies that \(U(0) = 0\), which implies that \(C = 0\).
So, the potential energy as a function of displacement is:
\[ U(x) = -\frac{k}{2}x^2 \]
This highlights that the potential energy decreases quadratically as the object moves away from the origin.

Integration of Force

The process of integrating force to find potential energy is essential in understanding energy dynamics. Integration allows us to reverse the differentiation process, which, in physical terms, lets us construct the energy landscape from the force interactions.
For our exercise, by integrating the negative of the force \(-kx\) with respect to \(x\), we derive the potential energy function \(U(x)\). This integration calculates the accumulated energy as the particle is displaced from the origin.
The integral we solve is:
\[ U(x) = \int -kx \, dx \]
Which results in:
\[ U(x) = -\frac{k}{2}x^2 + C \]
By applying the initial condition \(U(0) = 0\), we set the integration constant \(C\) to zero:
\[ U(x) = -\frac{k}{2}x^2 \]
The integration process not only provides the formula but also hints at the physical significance of potential energy. The negative sign suggests that energy decreases as the particle moves in the direction of the force. This example illustrates how integration bridges force and energy, making it crucial for understanding energy conservation in physical systems.