Problem 12 - Master Resource Book in JEE Main Physics

Question

A plate of mass \(m\), length \(b\) and breadth \(a\) is initially lying on a horizontal floor with length parallel to the floor and breadth perpendicular to the floor. The work done to erect it on its breadth is (a) \(m g\left[\frac{b}{2}\right]\) (b) \(m g\left[a+\frac{b}{2}\right]\) (c) \(m g\left[\frac{b-a}{2}\right]\) (d) \(m g\left[\frac{b+a}{2}\right]\)

Step-by-Step Solution

Answer

The correct answer is (a) \( mg \left[ \frac{b}{2} \right] \).

Step 1: Identify Initial and Final Positions

Understand that the plate is initially lying with its length parallel to the floor and its breadth perpendicular to it. This means its center of mass is at height 0. When the plate stands on its breadth, its center of mass rises to a height of \( \frac{b}{2} \).

Step 2: Calculate the Change in Height

The center of mass initially is at height 0 since the plate is laying flat. After the plate is raised to stand on its breadth, the center of mass moves to a height \( \frac{b}{2} \). Therefore, the change in height \( \Delta h \) is \( \frac{b}{2} \).

Step 3: Apply the Work Done Formula

The work done to raise an object is the change in gravitational potential energy, calculated by \( W = mgh \). Substitute the change in height, \( \Delta h = \frac{b}{2} \), into this formula to get \( W = mg \left( \frac{b}{2} \right) \).

Step 4: Verify with Given Options

Compare the calculated work done expression \( mg \left( \frac{b}{2} \right) \) with the provided options. It matches option (a) \( mg \left[ \frac{b}{2} \right] \).

Key Concepts

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. This concept is crucial when explaining how work can change the energy stored in an object. When you lift an object higher, you increase its gravitational potential energy.

For any object, the gravitational potential energy is determined by the equation:

• Gravitational Potential Energy (GPE) = mass \( m \) multiplied by gravitational acceleration \( g \) and the height \( h \) it reaches: \( GPE = mgh \).

In our exercise, the plate starts on the floor and gains potential energy as it is raised to stand on its breadth. The change in gravitational potential energy tells us how much work is done to lift it.

Center of Mass

The concept of center of mass is key to understanding how an object's mass is distributed in space. The center of mass is the point where we can assume the total mass of the object is concentrated. It simplifies calculations in physics when determining how the object behaves, especially in gravitational fields.

In our problem, the plate’s center of mass begins at the floor, making the initial height \(0\). When the plate stands on its edge, the center of mass shifts upward. Specifically, it moves to half the plate's length, \(\frac{b}{2}\), since the plate's mass is evenly distributed.

Recognizing the center of mass helps us compute the change in height accurately, which is fundamental for determining the work done.

Change in Height

The change in height is an essential element in calculating work done, especially in contexts involving gravitational potential energy. It refers to the vertical distance the center of mass moves when the object is repositioned.

In the given exercise, the plate initially lies flat, making the starting height of the center of mass zero. Once repositioned to stand on its breadth, the center of mass is raised to a height \(\frac{b}{2}\), signifying an overall change in height equal to \(\frac{b}{2}\).

Calculating this change accurately is crucial since it directly affects the calculation of work done using the formula:

• Work Done \( W = mg \Delta h \).

This equation shows that work is done against gravity when the center of mass is lifted, transforming the system's potential energy.