Problem 428 - JEE Physics

Question

A body of mass \(6 \mathrm{~kg}\) is under a force, which causes a displacement in it given by \(\mathrm{S}=\left[\left(2 \mathrm{t}^{3}\right) / 3\right](\mathrm{in} \mathrm{m}) .\) Find the work done by the force in first one seconds. (A) \(2 \mathrm{~J}\) (B) \(3.8 \mathrm{~J}\) (C) \(5.2 \mathrm{~J}\) (D) \(24 \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done by the force in the first one second is 12 J. The correct answer is (D).

Step 1: Find the velocity function

To find the velocity function, we need to differentiate the given displacement function with respect to time. Given, S(t) = (2t³) / 3 Differentiate S(t) with respect to t: v(t) = dS/dt = d(2t³/3)/dt = 2t²

Step 2: Find the acceleration function

To find the acceleration function, we need to differentiate the velocity function with respect to time. v(t) = 2t² Differentiate v(t) with respect to t: a(t) = dv/dt = d(2t²)/dt = 4t

Step 3: Find the force function

To find the force function, we should substitute the value of mass (m = 6 kg) and the acceleration function in F = m*a. a(t) = 4t F(t) = m*a(t) = 6*(4t) = 24t

Step 4: Find the work done during the first second

To find the work done during the first second, we need to integrate the force function over the time interval from 0 to 1 second. W = ∫(F(t)dt) from 0 to 1 W = ∫(24t*dt) from 0 to 1 Now, integrate the function: W = (12t²) | from 0 to 1 = 12(1)² - 12(0)² = 12 J So, the work done by the force in the first one second is 12 J. The correct answer is (D) 12 J.

Key Concepts

Kinematics

Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion. It typically involves quantities such as displacement, velocity, and acceleration. In this specific exercise, we're given a displacement function expressed as \( S = \frac{2t^3}{3} \), which describes how the position of a body changes over time.
To solve the problem, we first need to determine how this displacement changes over time, which directly leads us to velocity and then to acceleration, paving the way to understanding the motion of the body comprehensively.

Differentiation in Calculus

Differentiation is a crucial process in calculus that allows us to find the rate of change of a quantity. In physics, it's often used to determine how quantities like displacement and velocity change over time.
In the context of this exercise, differentiation helps us convert the displacement function \( S(t) = \frac{2t^3}{3} \) into a velocity function. We differentiate \( S(t) \) with respect to time \( t \), giving us a new function \( v(t) = dS/dt = 2t^2 \).
This velocity function \( v(t) \), representing the rate of change of displacement, is fundamental in calculating further motion properties. Differentiation further helps us ascertain the acceleration by differentiating the velocity function, which results in \( a(t) = dv/dt = 4t \), signifying the velocity's rate of change over time. Differentiation thus acts as the stepping stone toward completely understanding an object's motion.

Integration in Calculus

Integration is the reverse process of differentiation. It allows us to find a quantity from its rate of change, essentially adding up small increments to find the whole. In the context of physics, we use integration to calculate total quantities like work done by a force.
In this exercise, to find the work done, we have integrated the force over time. Given the force function \( F(t) = 24t \), we integrate this with respect to time over the interval from \( t = 0 \) to \( t = 1 \). This process sums up the small bits of work done over infinitesimal time intervals to get the total work done.
The integration steps give us \( W = \int_{0}^{1} 24t\, dt = 12t^2 \bigg|_{0}^{1} = 12 \times (1^2) - 12 \times (0^2) = 12 \) joules. This result shows how integration is used powerfully to calculate total quantities from moment-by-moment changes, indicative of the calculus principle named the Fundamental Theorem of Calculus.

Newton's Second Law

Newton's Second Law describes how the velocity of an object changes when subjected to an external force. It states that the force exerted on an object equals the mass of the object times its acceleration \((F = m \times a)\). This fundamental law of motion links acceleration and force, crucial for solving many physics problems.
In our exercise, once we determined the acceleration \( a(t) = 4t \) through differentiation, we applied Newton's Second Law using a mass of \( 6 \) kg. This resulted in the force function \( F(t) = m \times a(t) = 6 \times 4t = 24t \).
This force function is then used to compute the work done over time by integrating it, portraying how Newton's Second Law forms a foundation for dynamic physical analysis. By extending from fundamental principles like these, we connect seemingly simple math operations with real-world physical phenomena, making Newton's Second Law an integral part of understanding motion on a deeper level.