Problem 46 - Master Resource Book in JEE Main Physics

Question

A particle is displaced from a position \((2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})\) to another position \((3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})\) under the action of the force \((2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})\). The work done by the force is an arbitrary unit is (a) 8 (b) 10 (c) 12 (d) 16

Step-by-Step Solution

Answer

The work done was calculated as 6 units, but it does not match any of the given options.

Step 1: Identify Initial and Final Position Vectors

The initial position vector is \( \mathbf{r}_1 = 2\hat{\mathbf{i}} - \hat{\mathbf{j}} - \hat{\mathbf{k}} \). The final position vector is \( \mathbf{r}_2 = 3\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 2\hat{\mathbf{k}} \).

Step 2: Calculate the Displacement Vector

The displacement vector \( \Delta \mathbf{r} \) is calculated as follows:\[\Delta \mathbf{r} = \mathbf{r}_2 - \mathbf{r}_1 = (3\hat{\mathbf{i}} + 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}}) - (2\hat{\mathbf{i}} - \hat{\mathbf{j}} - \hat{\mathbf{k}})\]Simplify to get:\[\Delta \mathbf{r} = (3-2)\hat{\mathbf{i}} + (2+1)\hat{\mathbf{j}} + (-2+1)\hat{\mathbf{k}} = \hat{\mathbf{i}} + 3\hat{\mathbf{j}} - \hat{\mathbf{k}}\]

Step 3: Identify the Force Vector

The force vector \( \mathbf{F} \) is given by \( 2 \hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}} \).

Step 4: Calculate the Work Done

The work done is given by the dot product of the force vector and the displacement vector:\[W = \mathbf{F} \cdot \Delta \mathbf{r} = (2 \hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}}) \cdot (\hat{\mathbf{i}} + 3\hat{\mathbf{j}} - \hat{\mathbf{k}})\]Calculate the dot product:\[W = (2)(1) + (1)(3) + (-1)(-1) = 2 + 3 + 1 = 6\]

Step 5: Compare with Answer Choices

The calculated work done is 6 arbitrary units. None of the choices provided (8, 10, 12, 16) match the computed work done. Recheck the calculations for any errors.

Key Concepts

Displacement Vector

The displacement vector plays a crucial role in determining how far and in what direction an object moves. It is a vector quantity that represents the change in position of an object. To compute the displacement vector, we subtract the initial position vector from the final position vector.

In the problem, the initial position vector is given as \( \mathbf{r}_1 = 2\hat{\mathbf{i}} - \hat{\mathbf{j}} - \hat{\mathbf{k}} \). The final position vector is \( \mathbf{r}_2 = 3\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} - 2\hat{\mathbf{k}} \). The difference of these two vectors gives us the displacement:\[ \Delta \mathbf{r} = (3\hat{\mathbf{i}} + 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}}) - (2\hat{\mathbf{i}} - \hat{\mathbf{j}} - \hat{\mathbf{k}}) = \hat{\mathbf{i}} + 3\hat{\mathbf{j}} - \hat{\mathbf{k}} \]

This displacement vector \( \Delta \mathbf{r} \) shows us that the object has moved one unit in the direction of \( \hat{\mathbf{i}} \), three units in the direction of \( \hat{\mathbf{j}} \), and one unit backward in the direction of \( \hat{\mathbf{k}} \). Understanding the displacement vector helps us visualize the path taken by the object and is essential for calculating work done by a force.

Dot Product

The dot product, also known as the scalar product, is a method used in vector mathematics to combine two vectors and get a scalar outcome. This operation is pivotal when calculating the work done because work done by a force on an object is defined as the dot product of the force vector and the displacement vector.

The formula for calculating the dot product of two vectors \( \mathbf{A} = a_1\hat{\mathbf{i}} + a_2\hat{\mathbf{j}} + a_3\hat{\mathbf{k}} \) and \( \mathbf{B} = b_1\hat{\mathbf{i}} + b_2\hat{\mathbf{j}} + b_3\hat{\mathbf{k}} \) is:

• \( \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3 \)

In this exercise, using \( \mathbf{F} = 2 \hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}} \) and \( \Delta \mathbf{r} = \hat{\mathbf{i}} + 3\hat{\mathbf{j}} - \hat{\mathbf{k}} \), the dot product is calculated as follows:

• \( (2)(1) + (1)(3) + (-1)(-1) = 2 + 3 + 1 = 6 \)

This resulting scalar value of 6 indicates the magnitude of work done, demonstrating how effectively a force contributes to displacement along the action line of the force.

Force Vector

A force vector represents the force applied to an object in terms of its magnitude and direction. In this exercise, it is described as \( \mathbf{F} = 2 \hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}} \), signifying a force with components acting in the directions of the unit vectors \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \) and \( \hat{\mathbf{k}} \).

To understand the work done, we need to consider how this force applies along the displacement of the object. The force vector components are as follows:

• 2 units in the positive \( \hat{\mathbf{i}} \) direction
• 1 unit in the positive \( \hat{\mathbf{j}} \) direction
• -1 unit in the negative \( \hat{\mathbf{k}} \) direction

This force vector helps us quantify the interaction between the force applied and the movement of the object. When evaluating work done, it's important to see how much of the force effectively pushes the object along the direction of each component of its displacement. Understanding the vector nature of forces allows us to see the big picture of movement dynamics in physics.