Problem 83 - Master Resource Book in JEE Main Physics

Question

A small iron sphere is dropped from a great height. It attains its terminal velocity after having fallen \(32 \mathrm{~m}\). Then, it covers the rest of the path with terminal velocity only. The work done by air friction during the first \(32 \mathrm{~m}\) of fall is \(W_{1}\). The work done by air friction during the subsequent \(32 \mathrm{~m}\) fall is \(W_{2}\). Then \((a) W_{1}>W_{2}\) (b) \(W_{1}

Step-by-Step Solution

Answer

(a) \( W_{1} > W_{2} \)

Step 1: Analyze the Problem

We are given that a small iron sphere attains terminal velocity after falling 32 meters. Therefore, in the first 32 meters, it accelerates due to gravity but encounters increasing air friction until it balances the gravitational force, achieving terminal velocity thereafter.

Step 2: Understanding Forces in the First 32 m

In the first 32 m of fall, the forces acting on the sphere include gravitational force and air friction. Initially, air friction is less than the gravitational force, allowing acceleration. The work done by air friction, \(W_{1}\), is significant because it acts against the sphere as it accelerates to terminal velocity.

Step 3: Analysis for Subsequent 32 m

Once the terminal velocity is reached, the net force on the sphere is zero. The gravitational force is completely balanced by air friction. Therefore, the work done by air friction, \(W_{2}\), equals precisely the gravitational work over the 32 m path, as it counterbalances gravitational force entirely.

Step 4: Comparing Work Done in Both Segments

In the first 32 m, until terminal velocity is reached, the sphere spends energy to overcome the increasing friction and accelerate. In the subsequent 32 m, the work done by air friction, \(W_{2}\), remains constant at its maximum value, simply opposing gravity. Thus, \(W_{1} > W_{2}\), since acceleration requires additional work to overcome increasing friction.

Key Concepts

Air Friction

Air friction, also known as air resistance, plays a significant role in the motion of objects falling through the air. It is a type of friction that acts opposite to the object's direction of movement. As an object like the small iron sphere falls, it initially experiences less air friction.
This means it can accelerate downwards due to gravitational force. However, as its speed increases, so does the air friction, because the frictional force becomes stronger as the object moves faster.

* Air friction is a force counteracting the object's motion.
* It increases with the speed of the object.
* Eventually balances gravitational force at terminal velocity.

Once the sphere reaches a certain speed, called terminal velocity, the air friction matches the gravitational pull. At this point, there is no further acceleration and the object continues to fall at a constant speed.

Gravitational Force

Gravitational force is the force that pulls objects towards the center of the Earth. It is a constant force acting downwards on objects. For our iron sphere, gravity is the main force that starts off the acceleration during its fall.
This force remains the same, constantly pulling the sphere towards the Earth throughout its journey. Gravitational force is calculated with the formula:
\[F_g = mg\] where \( F_g \) is gravitational force, \( m \) is mass, and \( g \) is gravitational acceleration, usually approximated as \( 9.8 \text{ m/s}^2 \).

* Responsible for initial acceleration.
* Constantly acts downwards.
* Its effect is balanced by air friction at terminal velocity.

When the sphere reaches terminal velocity, gravitational force is fully counteracted by air friction, causing the sphere to continue its descent at a steady speed without further acceleration.

Work Done

In physics, "work done" is the measure of energy transfer when a force causes an object to move. It is essential to understand this concept to determine energy changes in different scenarios.
For the falling iron sphere, two significant segments during its journey need to be considered regarding work done:* **First 32 m:** Work done by air friction, \( W_1 \), involves the sphere accelerating to terminal velocity. Air friction increases gradually in this phase, requiring additional energy to overcome it.* **Subsequent 32 m:** Once terminal velocity is reached, work done by air friction, \( W_2 \), stays constant, simply opposing the gravitational pull.

The formula for work is:
\[ W = F \cdot d \cdot \cos\theta\]where \( W \) is work done, \( F \) is force, \( d \) is distance, and \( \theta \) is the angle between force and direction of motion. In our case, \( \cos\theta = 1 \) since the force due to air friction acts directly opposite to motion.
This notion explains why \( W_1 > W_2 \). During the first 32 m, the sphere does more energy-intensive work overcoming the rising air friction as it speeds up, whereas in the next 32 m, the work done by the friction opposes gravity steadily.