Problem 44 - Master Resource Book in JEE Main Physics

Question

A force, \(\mathbf{F}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}} \mathbf{N}\) displaces a particle through \(\mathbf{S}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{k}} \mathbf{m}\) in \(16 \mathrm{~s}\). The power developed by \(\mathbf{F}\) is (a) \(0.25 \mathrm{~J} \mathrm{~s}^{-1}\) (b) \(25 \mathrm{~J} \mathrm{~s}^{-1}\) (c) \(225 \mathrm{~J} \mathrm{~s}^{-1}\) (d) \(450 \mathrm{~J} \mathrm{~s}^{-1}\)

Step-by-Step Solution

Answer

The power developed is \(0.25 \text{ J/s}\)

Step 1: Calculate Work Done

The work done by the force \( \mathbf{F} \) on the particle is given by the dot product of the force and displacement vectors: \( W = \mathbf{F} \cdot \mathbf{S} \). First, substituting the given vectors, we have: \( \mathbf{F} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} \) and \( \mathbf{S} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{k}} \). Calculate the dot product: \( W = (2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}}) \cdot (2 \hat{\mathbf{i}} + 2 \hat{\mathbf{k}}) \).

Step 2: Solve for Dot Product

Perform the dot product operation: \( W = 2 \cdot 2 + 2 \cdot 0 + 0 \cdot 2 + 0 \cdot 2 = 4 \). Since the terms involving \( \hat{\mathbf{j}} \cdot \hat{\mathbf{k}} \) and \( \hat{\mathbf{k}} \cdot \hat{\mathbf{i}} \) are zero, they do not contribute to the dot product.

Step 3: Determine Power Developed

Power is defined as the work done per unit time. Use the formula \( P = \frac{W}{t} \), where \( W \) is the work and \( t \) is the time. Given that \( W = 4 \text{ J} \) and \( t = 16 \text{ s} \), calculate \( P = \frac{4}{16} = 0.25 \text{ J/s} \).

Key Concepts

Dot Product in Physics

The dot product is a fundamental operation in vector analysis and is essential for calculating work done by a force. It involves taking two vectors and returning a scalar quantity. In our context, the force \( \mathbf{F} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} \) and displacement \( \mathbf{S} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{k}} \) are the two vectors that need to be considered.

• Definition: The dot product \( \mathbf{A} \cdot \mathbf{B} \) is calculated as \( A_xB_x + A_yB_y + A_zB_z \) for vectors \( \mathbf{A} \) and \( \mathbf{B} \).
• Component Analysis: In our example, each vector has components along different unit vectors \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \).
• Relevant Calculation: Calculate the dot product as\[W = (2 \cdot 2) + (2 \cdot 0) + (0 \cdot 2) = 4\]This results in a scalar quantity indicating the work done.

The dot product plays a crucial role in finding how much of one vector lies in the direction of another, which is key in determining the actual contribution of a force to doing work.

Work Done by Force

Work done by a force is a key concept in physics. It refers to the energy transferred when a force acts upon an object to cause displacement. In simpler terms, it measures how much "effort" a force uses to move an object over a distance. For the calculation of work done, the dot product of force and displacement vectors is employed.

• Formula: \( W = \mathbf{F} \cdot \mathbf{S} \), where \( \mathbf{F} \) is the force vector, and \( \mathbf{S} \) is the displacement vector.
• Units: The work done is measured in joules (\( \text{J} \)).
• Context in Problem: For the vectors \( \mathbf{F} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} \) and \( \mathbf{S} = 2 \hat{\mathbf{i}} + 2 \hat{\mathbf{k}} \), the work done \( W \) is calculated using their dot product, giving us 4 J.

Understanding the concept of work done is crucial for solving problems related to energy transfer, as it indicates how effectively a force contributes to an object's movement.

Vector Analysis in Physics

Vector analysis is an essential tool in physics, allowing us to work with quantities that have both magnitude and direction, such as displacement and force. Knowing how to manipulate these vectors is crucial to solving many physics problems, like the one presented where we calculated how force creates work.

• Key Concepts:
  • Vectors: Entities with both magnitude and direction, typically represented in terms of unit vectors \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \).
  • Operations: In vector analysis, addition, subtraction, dot product, and cross product are crucial operations.
• Application in Given Problem: The problem involves force and displacement vectors, each plotted in a three-dimensional coordinate system with components along \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \). Each contributes differently based on its direction and magnitude.
• Result Interpretation: With vector analysis, we can determine how effective a force is in moving an object in a particular direction. This is exemplified by the work calculation through the dot product.

Vector analysis not only aids in calculations but also helps in visualizing the physical significance behind forces and movements in physics, making it a critical part of studying mechanics.