Problem 1 - Master Resource Book in JEE Main Physics

Question

Under the action of a force, a \(2 \mathrm{~kg}\) body moves such that its position \(x\) as a function of time \(t\) is given by \(x=t^{3} / 3\), where \(x\) is in metre and \(t\) in second. The work done by the force in the first two seconds is (a) \(1.6 \mathrm{~J}\) (b) \(16 \mathrm{~J}\) (c) \(160 \mathrm{~J}\) (d) \(1600 \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done by the force in the first two seconds is 64 J.

Step 1: Define the Work Done Formula

Work done by a force is given by the formula \[ W = \int F \cdot dx \] where \( F \) is the force and \( dx \) is the change in position. To find \( F \), we need to know the acceleration of the body.

Step 2: Determine Velocity and Acceleration

The velocity \( v \) is the derivative of the position function \( x(t) \) with respect to time \( t \). \[ v = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) = t^2 \]The acceleration \( a \) is then the derivative of velocity: \[ a = \frac{dv}{dt} = \frac{d}{dt}(t^2) = 2t \]

Step 3: Calculate the Force

Since force \( F \) is mass \( m \) times acceleration \( a \), we have: \[ F = ma = 2 \times 2t = 4t \]

Step 4: Integrate Force to Find Work Done

Substitute \( F = 4t \) into the work done formula:\[ W = \int_{0}^{2} 4t \cdot \, dx \] First, express \( dx \) in terms of \( dt \):\( dx = v dt = t^2 dt \).Thus,\[ W = \int_{0}^{2} 4t \cdot t^2 \, dt = \int_{0}^{2} 4t^3 \, dt \]

Step 5: Solve the Integration

Perform the integration:\[ W = 4 \int_{0}^{2} t^3 \, dt = 4 \left[ \frac{t^4}{4} \right]_{0}^{2} \]Simplify and calculate:\[ W = \left[ t^4 \right]_{0}^{2} = 2^4 - 0^4 = 16 \]Thus,\[ W = 4 \times 16 = 64 \text{ J} \]

Key Concepts

Kinematics and Motion

Kinematics is a branch of physics that describes the motion of objects without considering the forces that cause this motion. For any given motion, kinematics helps us determine positions, velocities, and accelerations of moving objects. In the provided exercise, the position of a body is described by a given function, \( x = \frac{t^3}{3} \), where \( x \) is measured in meters and \( t \) in seconds. This equation indicates how the body's position changes over time.
To analyze the motion using this equation, we need to determine both velocity and acceleration:

• Velocity: It's the rate of change of position with time. It is derived from the first derivative of the position function \( x(t) \). For this problem, the velocity \( v\) turns out to be \( t^2 \), showing how fast the object's position is changing every second.
• Acceleration: This is the rate of change of velocity with time, or simply the second derivative of the position function. Here, \( a = 2t \), expressing that acceleration increases linearly with time.

The relationship between these quantities helps us understand how the object moves under the influence of a force over a defined period.

Newton's Second Law of Motion

Newton's Second Law of Motion plays a critical role here by connecting force to the resulting motion of a body. It states that the force acting on an object is equal to the mass \( m \) of the object times its acceleration \( a \), expressed as \( F = ma \). This foundational principle provides the means to calculate the force when the mass and acceleration of an object are known.
In this exercise, the object's mass is given as \( 2 \, \mathrm{kg} \), and with an acceleration \( a = 2t \):

• Using \( F = ma \), we substitute in the values to find the force: \( F = 2 imes 2t = 4t \).
• This expression demonstrates that force evolves with time, aligned with changes in the object's acceleration.

This application of Newton's Second Law allows us to bridge the kinematic description of motion with the dynamic interactions causing it, forming a complete picture of how forces govern the movement of bodies.

Integration in Physics for Work Done

Integration is a powerful tool in physics, especially when determining work done by a force over a certain path. Work done by a force is defined by the integral \( W = \int F \cdot dx \), where \( F \) is the force and \( dx \) the infinitesimal displacement.
The exercise demonstrates how integration ties in with kinematics and dynamics to calculate the work done by a time-dependent force:

• First, the force \( F \) as a function of time was determined (\( F = 4t \)).
• The displacement \( dx \) was expressed in terms of \( dt \) using the velocity function, resulting in \( dx = t^2 \, dt \).
• Substitute these into the integral for work: \( W = \int_{0}^{2} 4t \, t^2 \, dt = \int_{0}^{2} 4t^3 \, dt \).
• Carrying out the integration, \( W = 4 \left[ \frac{t^4}{4} \right]_{0}^{2} \), simplifies to \( W = 4 \times 16 = 64 \text{ J} \).

The integral applied here sums up the continuous application of force over the body's displacement, giving us the total work done, which is crucial for understanding energy transfer in physical systems.