Problem 6 - Master Resource Book in JEE Main Physics

Question

A \(5 \mathrm{~kg}\) brick of \(20 \mathrm{~cm} \times 10 \mathrm{~cm} \times 8 \mathrm{~cm}\) dimensionless lying on the largest base. It is now made to stand with length vertical. If \(g=10 \mathrm{~ms}^{-2}\), then the amount of work done is (a) \(3 \mathrm{~J}\) (b) \(5 \mathrm{~J}\) (c) 7\(]\) (d) \(9 \mathrm{~J}\)

Step-by-Step Solution

Answer

3 J (option a).

Step 1: Find initial center of mass height

Initially, the brick is lying on its largest base, which is the base with dimensions \(20\, \text{cm} \times 10\, \text{cm}\). The height of the center of mass above the ground is half of the smaller height of the brick \(8\, \text{cm} = 0.08\, \text{m}\). Therefore, the initial height of the center of mass is \(0.04\, \text{m}\).

Step 2: Find final center of mass height

After being made to stand with the length vertical, the longest dimension \(20\, \text{cm} = 0.2\, \text{m}\) now becomes the height of the brick. The center of mass will be at half of this height, which is \(0.1\, \text{m}\).

Step 3: Calculate the change in height of the center of mass

The center of mass moves from an initial height of \(0.04\, \text{m}\) to a final height of \(0.1\, \text{m}\). Therefore, the change in height is \(0.1 - 0.04 = 0.06\, \text{m}\).

Step 4: Calculate the work done

The work done in raising the center of mass of the brick is calculated by the formula \( \text{Work} = mgh \), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the change in height.Substituting the values: \( m = 5\, \text{kg} \), \( g = 10\, \text{ms}^{-2} \), and \( h = 0.06\, \text{m} \), we get the work done as:\[ \text{Work} = 5\times 10\times 0.06 = 3\, \text{J} \].

Step 5: Choose the correct option

The calculated work done is \(3 \text{ J}\), which corresponds to option (a).

Key Concepts

Understanding Work Done

Work done is a key concept in physics that describes the energy transferred when a force is applied to move an object over a distance. It's not just about pushing or pulling; it's about how force causes displacement.

The formula for work done is: \[ \text{Work} = F \times d \times \cos(\theta) \]

where:

• \( F \) is the force applied.
• \( d \) is the displacement moved by the object in the direction of the force.
• \( \theta \) is the angle between the force and the displacement direction.

The unit of work done is the Joule (J). In our example, the force is the weight of the brick (mass \( m \times g \)), the displacement is the change in height, and the entirety of the force acts along the direction of the movement.

This means the work done in moving the brick's center of mass to a new height is entirely used in overcoming the gravitational force acting on it.

Role of Gravity

Gravity is the force that pulls objects towards the Earth. It's a natural force that's always acting on every object, giving weight to them. The weight of an object is calculated as the mass of the object multiplied by the gravitational acceleration, \( g \), which is approximately \( 9.81 \, \text{ms}^{-2} \) on Earth. For ease, it is sometimes rounded to \( 10 \, \text{ms}^{-2} \), as we have done in our exercise.

When you lift an object, you're working against gravity. Any time there is a change in height, as seen when the brick is moved from lying flat to standing vertically, work is done against this gravitational force.

The work done is directly proportional to the change in height—a concept clearly illustrated in the exercise problem. Grasping the role of gravity allows us to fully appreciate how natural forces affect everyday actions and simple physics problems alike.

Understanding Basic Mechanics

Mechanics is a branch of physics that deals with motion and the forces that affect motion. It encompasses concepts such as velocity, acceleration, force, and energy—including work and gravity. In this exercise, we are examining a very basic mechanical system: a brick being rotated from flat to upright, which involves a change in position and energy.

The shift of the brick from a flat position to a vertical standing position brings dynamics into the picture—since it involves a change in the state of rest relative to its center of mass and its interaction with the gravitational pull.

Understanding this allows one to see how mechanics applies to the process: how forces cause movement, how the energy associated with the brick changes, and how this specific work problem provides insight into broad mechanics principles. Breaking down such problems is the heart of studying mechanics: observing how forces interact and cause changes in motion.