Problem 14 - Master Resource Book in JEE Main Physics

Question

Power supplied to a particle of mass \(2 \mathrm{~kg}\) varies with time as \(P=t^{2} / 2\) watt, where \(t\) is in second. If velocity of particle at \(t=0\) is \(v=0\), the velocity of particle at \(t=2 \mathrm{~s}\) will be (a) \(1 \mathrm{~ms}^{-1}\) (b) \(4 \mathrm{~ms}^{-1}\) (c) \(2 \sqrt{\frac{2}{3}} \mathrm{~ms}^{-1}\) (d) \(2 \sqrt{2} \mathrm{~ms}^{-1}\)

Step-by-Step Solution

Answer

(c) \(2 \sqrt{\frac{2}{3}} \mathrm{~ms}^{-1}\)

Step 1: Understand the Problem

We need to find the velocity of a 2 kg particle at \(t = 2\text{ s}\), given that the power supplied to it varies with time as \(P = \frac{t^2}{2}\text{ watt}\), and its initial velocity at \(t = 0\) is \(v = 0\).

Step 2: Relate Power to Work and Kinetic Energy

Power \(P\) is the rate at which work is done, and work done is the change in kinetic energy. Therefore, if \(P = \frac{dW}{dt} = \frac{t^2}{2}\), then the incremental work done \(dW = \frac{t^2}{2} dt\).

Step 3: Express Work Done as Change in Kinetic Energy

The work-energy theorem states that the work done on an object is equal to the change in kinetic energy. Thus, \(W = \Delta KE = \frac{1}{2} mv^2 - \frac{1}{2} m(0)^2\). Here, mass \(m = 2\text{ kg}\) and initial velocity \(v = 0\text{ m/s}\).

Step 4: Integrate Power to Find Total Work Done

Integrate \(dW = \frac{t^2}{2} dt\) from \(t = 0\) to \(t = 2\) to find the total work done: \[ W = \int_0^2 \frac{t^2}{2} dt = \left[ \frac{t^3}{6} \right]_0^2 = \frac{8}{6} = \frac{4}{3} \text{ J} \]

Step 5: Calculate Resulting Velocity

Set \(W = \frac{4}{3} \text{ J}\) equal to the change in kinetic energy \(\frac{1}{2} \times 2 \times v^2\) and solve for \(v\):\[ \frac{4}{3} = \frac{1}{2} \times 2 \times v^2 \]\[ v^2 = \frac{4}{3} \]\[ v = \sqrt{\frac{4}{3}} = 2 \sqrt{\frac{1}{3}} = 2 \cdot \frac{1}{\sqrt{3}} \]Simplifying, we find \(v = 2 \cdot \sqrt{\frac{2}{3}}\).

Step 6: Verify with Options

Check which option matches the calculated velocity. The answer matches option (c) \(2 \sqrt{\frac{2}{3}} \text{ m/s}.\)

Key Concepts

Work-Energy Theorem

The Work-Energy Theorem is a fundamental principle in physics that connects the concepts of work and kinetic energy. It states that the work done on an object by the net force is equal to the change in its kinetic energy. This is mathematically expressed as: \[ W = \Delta KE = \frac{1}{2} mv^2 - \frac{1}{2} mv_0^2 \]where:- \( W \) is the work done,- \( \Delta KE \) is the change in kinetic energy,- \( m \) is the mass of the object,- \( v \) is the final velocity,- \( v_0 \) is the initial velocity.In this context, if we know the initial and final velocities of a particle, along with its mass, we can calculate the total work done on the particle through these changes. This theorem helps bridge the gap between the force applied and the resulting motion, allowing us to determine the kinetic energy at different time intervals. The theorem is particularly useful in situations where the power supplied to a particle changes over time, as it enables us to relate this variable power to a concrete change in the motion of the particle.

Integration in Physics

Integration is an essential tool in physics, especially when dealing with continuously varying quantities over time. When power, which is the rate of doing work, changes with time, we use integration to find the total work done over a period. In our exercise, power \( P \) is given by \( P = \frac{t^2}{2} \) watts. To compute the work done from time \( t = 0 \) to \( t = 2 \) seconds, we integrate the power function over this interval:\[ W = \int_{0}^{2} \frac{t^2}{2} \,dt \]This results in the total work done being \( \frac{4}{3} \) Joules, showing how integration can give us a clear picture of the total impact over time when dealing with changing quantities. Not only does this provide us with the work done, but it also bridges the gap between abstract mathematical concepts and their practical application in physics. Through integration, we see the cumulative effects in action as they relate to an object's motion and energy changes.

Particle Motion under Variable Power

When a particle is subjected to variable power, its motion can become quite complex as the acceleration and velocity are not constant but change over time. In these situations, understanding how velocity evolves requires a systematic approach using the relationship between power, work, and kinetic energy. In the given problem, the power supplied to the particle changes with time as \( P = \frac{t^2}{2} \), which suggests that different amounts of energy are imparted to the particle at different times. To determine how this affects the particle's motion, we must:- Calculate the work done by integrating the power function over the given time period.- Use the Work-Energy Theorem to relate this work to changes in kinetic energy.- Solve for the particle's velocity at the final time by setting the computed work equal to the change in kinetic energy, and solve for the velocity term.Thus, through the combined application of these principles, we find that the velocity of the particle at \( t = 2 \) seconds satisfies \( v = 2 \sqrt{\frac{2}{3}} \text{ m/s} \). This process showcases the intricacies of motion under variable conditions and highlights the analytic techniques necessary for accurate predictions in physics.