Problem 10 - Master Resource Book in JEE Main Physics

Question

A body of mass \(3 \mathrm{~kg}\) is under a force which causes a displacement is given by \(s=\frac{t^{3}}{3}(\) in \(\mathrm{m})\). Find the work done by the force in first 2 seconds. (a) 2\(]\) (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 is 24 J.

Step 1: Understand the Problem

To find the work done by the force, we need to use the formula for work done, which is given by the integral of the force over the displacement. Since we have the displacement as a function of time, we must first find the force acting on the body.

Step 2: Determine the Velocity

The displacement function is given by \(s = \frac{t^3}{3}\). To find the velocity, differentiate the displacement with respect to time: \(v = \frac{ds}{dt} = t^2\).

Step 3: Determine the Acceleration

Differentiate the velocity with respect to time to find the acceleration: \(a = \frac{dv}{dt} = 2t\).

Step 4: Calculate the Force

Using Newton's second law, \(F = ma\), where \(m = 3 \mathrm{~kg}\) and \(a = 2t\). Thus, the force is \(F = 3 \cdot 2t = 6t\).

Step 5: Calculate Work Done

Work done, \(W = \int_{t=0}^{t=2} F \cdot v \, dt\). Substitute \(F = 6t\) and \(v = t^2\), we get \(W = \int_{0}^{2} 6t \cdot t^2 \, dt = \int_{0}^{2} 6t^3 \, dt\).

Step 6: Evaluate the Integral

Evaluate \(\int_{0}^{2} 6t^3 \, dt = 6 \int_{0}^{2} t^3 \, dt\). Find the antiderivative: \(t^3\) integrates to \(\frac{t^4}{4}\), so we have \(6 \cdot \left[\frac{t^4}{4}\right]_{0}^{2}\) which evaluates to \(6 \cdot \left( \frac{16}{4} - 0 \right) = 24 \mathrm{~J}\).

Key Concepts

Newton's Second Law

Newton's Second Law is a fundamental principle in physics that links the concepts of force, mass, and acceleration. This law is usually expressed by the formula:

• \( F = ma \)

In this equation, \( F \) represents the force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration experienced by the object. The law essentially states that the force acting on an object is equal to the mass of that object multiplied by its acceleration.
This principle is often applied when calculating the force required to cause an object to move in a certain way. In the context of our exercise, we first derive the acceleration of the object from its velocity function to find the force acting on it. In practice, Newton's Second Law allows us to analyze motions and predict how objects will move when subjected to different forces. Understanding this law provides the foundation for solving many physics problems related to motion.

Displacement Function

A displacement function describes how the position of an object changes over time. It is an essential concept in understanding motion and finding related quantities such as velocity and acceleration. In our problem, the displacement function is given as:

• \( s = \frac{t^3}{3} \)

This equation indicates that the position \( s \) of the body depends on time \( t \), and it shows a cubic relationship. To understand motion thoroughly, one must be able to differentiate the displacement function.
Differentiating the displacement function with respect to time helps us find the velocity function. Velocity indicates the rate of change of displacement, and in our example, the differentiation results in:

• \( v = t^2 \)

This velocity function is crucial in further calculations, such as finding acceleration and eventually determining the work done by the force.

Velocity and Acceleration

Velocity and acceleration are closely connected with how an object moves. Velocity is the speed and direction of an object's motion, and it is calculated as the derivative of the displacement function. In our exercise, the velocity is:

• \( v = t^2 \)

This showcases that velocity changes as time progresses at a quadratic rate.Acceleration, on the other hand, is the rate of change of velocity over time. It's found by differentiating the velocity function:

• \( a = \frac{dv}{dt} = 2t \)

Here, the acceleration turns out to be directly proportional to time, \( t \). Comprehending these functions is vital because they bridge the gap between force and motion.
Understanding how velocity and acceleration relate to force—through Newton's Second Law—is key to calculating work done in physics, especially in situations where force isn't constant. Therefore, mastering these concepts is critical for solving complex motion problems efficiently.