Problem 39 - Master Resource Book in JEE Main Physics

Question

A body of mass \(0.5 \mathrm{~kg}\) travels in a straight line with velocity \(v=a x^{3 / 2}\) where \(a=5 \mathrm{~m}^{-1 / 2} \mathrm{~s}^{-1}\). The work done by the net force during its displacement from \(x=0\) to \(x=2 \mathrm{~m}\) is \(\quad\) [NCERT Exemplar] (a) \(1.5 \mathrm{~J}\) (b) \(50 \mathrm{~J}\) (c) \(10 \mathrm{j}\) (d) \(100 \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done is 50 J (option b).

Step 1: Understanding the Problem

We are given a velocity expression as a function of position, \(v = a x^{3/2}\), where \(a = 5 \, \text{m}^{-1/2}\, \text{s}^{-1}\). We need to find the work done by the net force as the body moves from \(x=0\) to \(x=2\) meters.

Step 2: Calculate the Acceleration

First, find the expression for acceleration \(a(x)\). We start by expressing velocity as \(v = \frac{dx}{dt}\). Differentiating with respect to \(t\) gives us acceleration: \(a(x) = \frac{dv}{dt}\). Since \(v = a x^{3/2}\), and using \(\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}\), we have \(a(x) = \frac{d(a x^{3/2})}{dx} \cdot v = \frac{3}{2}a x^{1/2} \cdot a x^{3/2} = \frac{3}{2}a^2 x^2\).

Step 3: Use the Work-Energy Theorem

The work done can be found using \(W = \int F dx\) where \(F = ma\). So, \(F = m \cdot \frac{3}{2}a^2 x^2\). Therefore, \(W = \int_{0}^{2} \frac{3}{2}m a^2 x^2 \, dx\). Substitute \(m = 0.5 \, \text{kg}\) and \(a = 5 \, \text{m}^{-1/2} \text{s}^{-1}\).

Step 4: Calculate the Work Done

Evaluating the integral gives \(W = \frac{3}{2}(0.5)(5^2) \int_{0}^{2} x^2 \, dx\). The definite integral \(\int_{0}^{2} x^2 \, dx\) evaluates to \(\left[\frac{x^3}{3}\right]_{0}^{2} = \frac{8}{3}\). Substituting gives \(W = \frac{3}{2} \times 0.5 \times 25 \times \frac{8}{3} = 50 \, \text{J}\).

Step 5: Verify the Result

Review steps and ensure calculations are correct. The expression for acceleration and force have been used correctly, and the integral was evaluated accurately. Since the calculations are consistent, the work done is verified as \(50 \, \text{J}\).

Key Concepts

Kinematics

Kinematics is the branch of physics that focuses on the motion of objects without considering the forces that cause the motion. It deals with parameters like velocity, acceleration, displacement, and time.
In this exercise, the object's velocity is given as a function of position: \(v = a x^{3/2}\). This equation provides a relationship between the speed of the body and its position along a straight path.

• **Velocity** is the rate at which an object's position changes. It is a vector, meaning it has both magnitude and direction.
• **Acceleration** is the rate of change of velocity with respect to time. For the given velocity function, the acceleration as a function of position is found by taking the derivative with respect to \(x\).

Understanding these concepts in kinematics allows us to set up equations needed to later calculate forces and work done as the object moves.

Integral Calculus

Integral calculus plays a crucial role in this exercise, especially in calculating the work done by the net force. Integrals allow us to sum up infinitely small quantities over a range, which is essential for determining quantities like work.

• In this problem, we use the integral \( W = \int F \, dx \) to calculate work. Here, \(F\) is the force acting on the object as a function of position \(x\).
• To find \(F\), we needed the acceleration \(a(x)\), which was derived from the velocity function using differentiation. The force then becomes \( F = m \cdot a(x) \).
• Finally, the work is computed by evaluating the integral of the force over the displacement from \(x = 0\) to \(x = 2\).

This approach demonstrates how integral calculus is used to compute real-world quantities involving continuous change and accumulation, crucial for solving complex problems in physics.

Forces and Motion

Forces and motion are intimately connected in physics and are a critical element of the work-energy theorem. The work-energy theorem tells us that the work done by the net force on an object is equal to the change in its kinetic energy.

• **Force** is a push or pull on an object that can cause it to accelerate. In this problem, it's related to the mass and acceleration of the object via Newton's second law, \( F = ma \).
• **Motion** refers to the change in position of the object over time. Here, we consider motion along a straight line.

The solution involves calculating the force based on the mass \(0.5 \, \text{kg}\) and the derived acceleration function \(a(x)\), followed by using integral calculus to integrate \(F\) over the distance \(0\) to \(2 \, \text{m}\).
This step links how kinematics, calculus, and dynamics converge to determine the work done via the work-energy theorem.