Problem 447 - JEE Physics

Question

A force \(F=A y^{2}+B y+C\) acts on a body in the \(y\) -direction. The work done by this force during a displacement from \(\mathrm{y}=-\mathrm{a}\) to \(\mathrm{y}=\mathrm{a}\) is (A) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]\) (B) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+2 \mathrm{ca}\) (C) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+\left[\left\\{\mathrm{Ba}^{2}\right\\} / 2\right]+\mathrm{ca}\) (D) None of these.

Step-by-Step Solution

Answer

The short answer for the work done is: (B) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+2 \mathrm{ca}\)

Step 1: Identify the given information

The force acting on the body in the y-direction is given by F = Ay² + By + C, and we need to find the work done during the displacement from y = -a to y = a.

Step 2: Write the formula for work done

The work done (W) by a force acting in the y-direction can be calculated by integrating the force F with respect to y over the given interval [-a, a]: \[W = \int_{-a}^{a} F(y) dy\]

Step 3: Substitute the expression for F

Replace F in the integral expression with Ay² + By + C: \[W = \int_{-a}^a (Ay^2 + By + C) dy\]

Step 4: Integrate the expression

Now, integrate each term with respect to y: \[W = \int_{-a}^a Ay^2 dy + \int_{-a}^a By dy + \int_{-a}^a C dy\] \[W = A\int_{-a}^a y^2 dy + B\int_{-a}^a y dy + C\int_{-a}^a dy\] For each integral: \[\int y^2 dy = \frac{1}{3} y^3 + D_1\] \[\int y dy = \frac{1}{2} y^2 + D_2\] \[\int dy = y + D_3\] Integration constants D1, D2, and D3 are not necessary, so the results are: \[W = A\left[\frac{1}{3} y^3\right]_{-a}^a + B\left[\frac{1}{2} y^2\right]_{-a}^a + C[y]_{-a}^a\]

Step 5: Evaluate the integral at the endpoints

Now, evaluate the expression at the endpoints y = -a and y = a: \[W = A\left(\frac{1}{3}(a^3) - \frac{1}{3}(-a)^3\right) + B\left(\frac{1}{2}(a^2) - \frac{1}{2}(-a)^2\right) + C(a - (-a))\]

Step 6: Simplify the expression

Simplify the expression for the work done: \[W = A\left(\frac{2a^3}{3}\right) + B\left(\frac{a^2}{2} - \frac{a^2}{2}\right) + C(2a)\] \[W = \frac{2Aa^3}{3} + 0 + 2Ca\]

Step 7: Choose the correct answer

The simplified expression for the work done is: \[W = \frac{2Aa^3}{3} + 2Ca\] Comparing this with the given choices, we find that the correct answer is (B) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+2 \mathrm{ca}\).

Key Concepts

Understanding Integration in Physics

In physics, integration is a crucial mathematical tool used to determine quantities like work done by a force over a specific path or interval. When a force is variable, as in the given problem, integration helps us sum up the small contributions of the force over the path of displacement.

In this exercise, the function describing the force is polynomial, given as \( F(y) = Ay^2 + By + C \). To find the work done by this force from \( y = -a \) to \( y = a \), we use integration. This process involves substituting the function into the integral, as shown in the solution steps, and computing the definite integral. The result provides us with the total work done.

Integration breaks down the continuous problem into manageable parts and adds them back together to give us a comprehensive solution. It's akin to adding a series of very thin slices to determine the entire area under the force curve along the path of motion.

Polynomial Force and Its Application

A polynomial force is an expression where the force is defined as a polynomial function of a variable, in this case, the variable \( y \). The given force \( F(y) = Ay^2 + By + C \) is a polynomial of degree 2. Each term represents a component of the force that depends on \( y \) differently.

• The first term \( Ay^2 \) indicates a quadratic dependence on \( y \), meaning it increases or decreases with the square of \( y \).
• The second term \( By \) suggests a linear relationship, where the force changes directly with \( y \).
• The final term, \( C \), is a constant, contributing uniformly along the path, unaffected by \( y \).

Understanding the nature of polynomial forces is important as they frequently occur in physics, particularly in complex systems where the force might vary with position, not uniformly.

Solving for work using such a function requires breaking down each component, as seen in the integral setup, integrating each term separately, and then combining them to evaluate the work done.

Physics Problem Solving Techniques

Problem-solving in physics often involves breaking down complex scenarios into simpler, more manageable parts. This technique is well-illustrated in solving for the work done by a polynomial force. Here is a general approach applied:

• **Identify the problem:** Understand what you're asked to find or prove. In this case, it's the work done.
• **Mathematical Representation:** Translate the physical scenario into mathematical terms. Write down the force function and the limits of displacement.
• **Use appropriate math tools:** Integration is the tool of choice for accumulating the work done by a varying force over a distance.
• **Solve step-by-step:** Integrate each term of the polynomial, evaluate at the boundaries, and combine the results.

This structured problem-solving process helps in tackling real-world physics problems effectively by maintaining focus, ensuring completeness, and systematically achieving the solution required.

Y-Direction Displacement and Its Influence

The displacement in the y-direction plays a pivotal role in this physics problem, dictating the limits over which the work is calculated. Displacement affects how much work is done because work is essentially a product of force and displacement. Here, the displacement is between \( y = -a \) and \( y = a \).

• The y-direction specifies one axis along which everything happens. In physics, knowing the direction is vital as it influences the integration limits and the behavior of each force component.
• This directional approach simplifies multi-dimensional problems by allowing us to focus on changes along a single line of action.
• The y-direction displacement here helps determine the symmetry in the problem, which simplifies calculations – noticed in the cancellation of like terms when evaluating the integrals.

Understanding the influence of directional displacement is essential for predicting outcomes, such as whether or not some forces will ultimately cancel out, simplifying calculations in physical problems.