Problem 61 - Master Resource Book in JEE Main Physics

Question

A point of application of a force \(\mathbf{F}=5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) is moved from \(\mathbf{r}_{1}=2 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) to \(\mathbf{r}_{2}=-5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \mathbf{k}\) the work done is (a) \(-17\) units (b) \(-22\) units (c) 33 units (d) \(-33\) units

Step-by-Step Solution

Answer

The work done is -17 units. The correct option is (a).

Step 1: Understand the Problem

The problem involves finding the work done by a force as an object moves from point \( \mathbf{r}_1 \) to point \( \mathbf{r}_2 \). The work done by a force \( \mathbf{F} \) when the object moves from point \( \mathbf{r}_1 \) to point \( \mathbf{r}_2 \) is given by the formula \( W = \mathbf{F} \cdot \mathbf{d} \), where \( \mathbf{d} \) is the displacement vector.

Step 2: Calculate the Displacement Vector

The displacement vector \( \mathbf{d} \) is the difference between the final position \( \mathbf{r}_2 \) and the initial position \( \mathbf{r}_1 \). So, \( \mathbf{d} = \mathbf{r}_2 - \mathbf{r}_1 = (-5 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}) - (2 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}}) \). Calculating this gives \( \mathbf{d} = (-7 \hat{\mathbf{i}} - 5 \hat{\mathbf{j}} - 1 \hat{\mathbf{k}}) \).

Step 3: Compute the Work Done

The work done \( W \) is calculated using the dot product formula: \( W = \mathbf{F} \cdot \mathbf{d} = (5 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) \cdot (-7 \hat{\mathbf{i}} - 5 \hat{\mathbf{j}} - 1 \hat{\mathbf{k}}) \). Compute the dot product by multiplying the corresponding components and adding them: \( W = (5)(-7) + (-4)(-5) + (2)(-1) = -35 + 20 - 2 = -17 \).

Step 4: Select the Correct Answer

Based on the calculation, the work done is \(-17\) units. Thus, the correct answer is option (a) \(-17\) units.

Key Concepts

Displacement Vector

A displacement vector is all about showing how an object has moved from one place to another in a straight line. It's kind of like drawing an arrow from where you start to where you finish. In our problem, we need to find the displacement vector because it tells us how the point of application of the force has moved.
We have two position vectors here, \(\mathbf{r}_1\) and \(\mathbf{r}_2\), and we find the displacement \(\mathbf{d}\) by taking the vector subtraction \(\mathbf{r}_2 - \mathbf{r}_1\). This subtraction lets us see the change in each direction (i, j, k). Remember:
\[ \mathbf{d} = \mathbf{r}_2 - \mathbf{r}_1 = (-5 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}) - (2 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}}) = -7 \hat{\mathbf{i}} - 5 \hat{\mathbf{j}} - 1 \hat{\mathbf{k}} \]
This vector \(\mathbf{d} = -7 \hat{\mathbf{i}} - 5 \hat{\mathbf{j}} - 1 \hat{\mathbf{k}}\) shows us how far and in what direction the movement has taken place.

Dot Product

The dot product is a special way of multiplying two vectors that gives us a scalar (a single number) rather than another vector. It's very useful in physics problems like this, where we want to find how much of one vector is going in the direction of another. In our exercise, we use the dot product to calculate work done by a force, which is a core concept.
The formula for the dot product is straightforward:
\[ \mathbf{F} \cdot \mathbf{d} = F_i \cdot d_i + F_j \cdot d_j + F_k \cdot d_k \]
For our vectors, \(\mathbf{F} = 5 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}\) and \(\mathbf{d} = -7 \hat{\mathbf{i}} - 5 \hat{\mathbf{j}} - 1 \hat{\mathbf{k}}\), the dot product calculation looks like this:
\[ W = (5)(-7) + (-4)(-5) + (2)(-1) = -35 + 20 - 2 = -17 \]
This gives us the work done, showing how the dissected vector components contribute to the task load done by the force in its direction of movement.

Vector Subtraction

Vector subtraction is an operation that helps us find the difference between two vectors, which we earlier used to find the displacement vector. When vectors represent positions, their subtraction gives a new vector that shows the movement from one position to another.
Here's a simple reminder of how vector subtraction works:
1. Subtract the corresponding components.
2. For instance, for vector A = \(2 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}}\) and vector B = \(-5 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}}\),\[ \mathbf{d} = \mathbf{B} - \mathbf{A} = (-5 - 2) \hat{\mathbf{i}} + (2 - 7) \hat{\mathbf{j}} + (3 - 4) \hat{\mathbf{k}} \]
This results in\[ \mathbf{d} = -7 \hat{\mathbf{i}} - 5 \hat{\mathbf{j}} - 1 \hat{\mathbf{k}} \]

Physics Problem Solving

Solving physics problems can sometimes feel like puzzle-solving. It involves breaking down the problem into manageable parts and using known equations and concepts. Here's a simple framework that may help you tackle similar problems:

• Understand the problem: Carefully read the question to identify what is being asked.
• Identify key concepts: Recognize the physics concepts involved like force, work, displacement.
• Write down known formulas: For work done by force, use \( W = \mathbf{F} \cdot \mathbf{d} \).
• Calculate step by step: Break down the calculations into steps like finding displacement and then using the dot product.
• Check your answer: Compare your work with possible answers or verify by recalculating.

Following these steps can help you master physics problem solving one concept at a time.