Problem 42 - Master Resource Book in JEE Main Physics

Question

A body of mass \(m\) is accelerated uniformly from rest to a speed \(v\) in a time \(T\). The instantaneous power delivered to the body as a function of time is given by [AIEEE 2008] (a) \(\frac{1}{2} \frac{m v^{2}}{T^{2}} t^{2}\) (b) \(\frac{1}{2} \frac{m v^{2}}{T^{2}} t\) (c) \(\frac{m v^{2}}{\tau^{2}} t^{2}\) (d) \(\frac{m v^{2}}{T^{2}} t\)

Step-by-Step Solution

Answer

The correct answer is (d): \( \frac{mv^2}{T^2} t \).

Step 1: Determine the Expression for Acceleration

Since the body starts from rest and reaches a velocity \( v \) in time \( T \), the acceleration \( a \) is given by the formula for uniform acceleration. Using the formula \( v = u + at \), where \( u \) is the initial velocity (0 in this case), we have \( v = aT \). Thus, \( a = \frac{v}{T} \).

Step 2: Determine the Expression for Force

Using Newton's second law, the force \( F \) applied to the body is the product of mass \( m \) and acceleration \( a \). So, \( F = ma = m \left( \frac{v}{T} \right) = \frac{mv}{T} \).

Step 3: Find the Expression for Velocity as a Function of Time

As the body accelerates uniformly from rest, the velocity at any time \( t \) is given by \( v(t) = at = \frac{v}{T}t \).

Step 4: Define Instantaneous Power

Instantaneous power \( P \) is defined as the rate at which work is done, or \( P = Fv \). Substitute the expressions for force and velocity: \( P = \left( \frac{mv}{T} \right) \left( \frac{v}{T} t \right) = \frac{mv^2}{T^2} t \).

Step 5: Match with Answer Choices

Looking at the derived expression \( P = \frac{mv^2}{T^2} t \), we compare with the given options and find that option (d) matches our derived formula.

Key Concepts

Uniform Acceleration

Uniform acceleration occurs when an object's velocity changes at a constant rate over time. This means that the acceleration, which is the rate of change of velocity, is the same throughout the motion. When a body starts from rest and reaches a velocity \( v \) in time \( T \), it is under uniform acceleration. The formula used to express this situation is \( v = u + at \), where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

Since the body starts from rest, \( u = 0 \), so the formula becomes \( v = at \). This can be rearranged to find the acceleration \( a \) as \( a = \frac{v}{T} \). This tells us how quickly the object's speed is increasing during the time of motion. Understanding uniform acceleration is crucial in solving problems regarding motion, forces, and energy, especially in physics problems as seen in the referenced exercise.

Newton's Second Law

Newton's Second Law of Motion provides a relationship between an object's mass, the acceleration it experiences, and the force applied to it. The law is expressed by the equation \( F = ma \). This formula implies that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

In the context of the exercise, once we know the acceleration \( a \) from the uniform acceleration calculation, we can determine the force applied to the object. If a body with mass \( m \) accelerates from rest to a velocity \( v \) over time \( T \), then the acceleration is \( a = \frac{v}{T} \). Therefore, the force acting on the body is \( F = m \left( \frac{v}{T} \right) = \frac{mv}{T} \). This equation helps us understand how force is derived and applied under conditions of uniform motion, and it sets the stage for calculating energy-related quantities like power.

Instantaneous Power Calculation

Instantaneous power is the power delivered to an object at a specific moment in time. It represents the rate at which work is done. In physics, power can be calculated using the formula \( P = Fv \), where \( F \) is the force applied and \( v \) is the velocity of the object at that moment.

In the given exercise, the force was calculated using Newton's second law as \( \frac{mv}{T} \). The velocity of the body at a time \( t \) is \( \frac{v}{T}t \), derived from uniform acceleration principles. Substituting these values into the power formula gives \( P = \left( \frac{mv}{T} \right) \left( \frac{v}{T} t \right) = \frac{mv^2}{T^2} t \).

This equation shows that the instantaneous power increases linearly with time, assuming constant force and acceleration. Understanding instantaneous power helps in analyzing how quickly energy is being transferred in a system, which is particularly useful in dynamics and engineering applications.