Problem 13 - Master Resource Book in JEE Main Physics

Question

A body of mass \(3 \mathrm{~kg}\) is under a force which causes a displacement in it, given by \(s=t^{2} / 3\) (in m). Find the work done by the force in \(2 \mathrm{~s}\) (a) \(2 \mathrm{~J}\) (b) \(3.8 \mathrm{~J}\) (c) \(5.2 \mathrm{~J}\) (d) \(2.6 \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done is approximately 2.67 J, closest to option (d) 2.6 J.

Step 1: Understand the Formula for Work

The work done by a force is given by the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement vectors. Given no information on direction, assume \( \theta = 0 \), meaning \( \cos(\theta) = 1 \). Thus, \( W = F \cdot d \).

Step 2: Find the Displacement Formula

We are given the displacement as a function of time: \( s = \frac{t^2}{3} \). This represents the displacement \( s \) after a time \( t \).

Step 3: Calculate Displacement at 2 Seconds

Substitute \( t = 2 \) into \( s = \frac{t^2}{3} \) to find the displacement after 2 seconds: \[ s = \frac{(2)^2}{3} = \frac{4}{3} \text{ meters} \].

Step 4: Determine the Force Using Newton's Second Law

According to Newton’s second law, \( F = m \cdot a \). We need to find the acceleration \( a \). First, differentiate the displacement function \( s = \frac{t^2}{3} \) with respect to time to find velocity, and differentiate again to find acceleration.

Step 5: Calculate Velocity and Acceleration

The velocity \( v = \frac{ds}{dt} = \frac{2t}{3} \). Differentiate to find acceleration: \( a = \frac{dv}{dt} = \frac{d}{dt}(\frac{2t}{3}) = \frac{2}{3} \text{ m/s}^2 \).

Step 6: Calculate the Force

Using \( F = m \cdot a \), substitute \( m = 3 \text{ kg} \) and \( a = \frac{2}{3} \text{ m/s}^2 \), giving: \( F = 3 \cdot \frac{2}{3} = 2 \text{ N} \).

Step 7: Calculate the Work Done

Now we find the work done by using \( W = F \cdot d \). With \( F = 2 \text{ N} \) and \( d = \frac{4}{3} \text{ m} \), \( W = 2 \cdot \frac{4}{3} = \frac{8}{3} \text{ J} \). Simplifying gives approximately \( 2.67 \text{ J} \).

Step 8: Compare to Options

The calculated work is approximately \( 2.67 \text{ J} \). The closest numerical option provided is (d) \( 2.6 \text{ J} \).

Key Concepts

Newton's Second Law

Newton's Second Law is a fundamental principle in physics that relates the force acting on an object to its mass and acceleration. It is expressed with the formula \( F = m \cdot a \), where \( F \) represents force, \( m \) is the mass, and \( a \) stands for acceleration. This law tells us that the greater the mass of an object, the more force is needed to change its state of motion. Simultaneously, increasing the force applied to an object increases its acceleration. Understanding this law is crucial in calculating work done by a force.Newton’s Second Law clarifies how:

• The amount of force affects an object’s acceleration.
• Mass is a measure of how much matter is in an object, affecting how much force is needed to achieve the same acceleration as a less massive object.
• The direction of the force impacts the direction of acceleration.

In practical problems, such as calculating the work done by a force on a moving object, knowing the mass and calculating the acceleration provides the force value necessary to compute work.

Displacement

Displacement refers to the overall change in position of an object from its initial position to its final position, measured in a straight line. It is a vector quantity, which means it has both magnitude and direction. Displacement is different from distance, which measures the total path length traveled; displacement focuses only on the initial and final positions.When dealing with physics problems:

• Displacement can be positive or negative, indicating direction.
• In the context of a force, displacement is the distance over which the force acts.
• Information on the path or route taken between the start and end is not necessary.

Given the formula \( s = \frac{t^2}{3} \) from the problem, displacement is calculated by substituting the time variable, enabling you to determine how far the object has moved after a specific period.

Acceleration

Acceleration is defined as the rate of change of velocity over time. It is a vector quantity, meaning it has both a magnitude and a specific direction. In everyday terms, acceleration can be thought of as how quickly an object speeds up, slows down, or changes direction.Key aspects of acceleration include:

• A change in speed or direction results in acceleration.
• Acceleration can be negative, implying the object is slowing down (also called deceleration).
• It is determined by differentiating the velocity with respect to time.

In the given exercise, the acceleration is derived by differentiating the velocity, which itself is found by differentiating the displacement with respect to time. This step-by-step differentiation leads to the equation \( a = \frac{2}{3} \text{ m/s}^2 \), giving insight into how quickly the object is speeding up under the constant force applied.