Problem 67 - Master Resource Book in JEE Main Physics

Question

An object moves along a straight line path from \(P\) to Q under the action of a force \((4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \mathrm{N}\). If the coordinates of \(P\) and \(Q\) in metres are \((3,3,-1)\) and \((2,-1,4)\) respectively, then the work done by the force is (a) \(+23 \mathrm{~J}\) (b) \(-23 \mathrm{~J}\) (c) \(1015 \mathrm{~J}\) (d) \(\sqrt{35}(4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \mathrm{J}\)

Step-by-Step Solution

Answer

The work done by the force is +23 J (option a).

Step 1: Calculate the displacement vector

The displacement vector \( \mathbf{d} \) is the vector from point \( P \) to point \( Q \). It can be calculated as \( \mathbf{d} = (2-3) \hat{\mathbf{i}} + (-1-3) \hat{\mathbf{j}} + (4+1) \hat{\mathbf{k}} = -1 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} + 5 \hat{\mathbf{k}} \).

Step 2: Use the dot product to find work done

Work done \( W \) by the force is given by the dot product of the force vector \( \mathbf{F} = 4 \hat{\mathbf{i}} - 3 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}} \) and the displacement vector \( \mathbf{d} \). Thus, \( W = \mathbf{F} \cdot \mathbf{d} = (4)(-1) + (-3)(-4) + (3)(5) \).

Step 3: Calculate the individual products in the dot product

Calculate each term: \( (4)(-1) = -4 \), \( (-3)(-4) = 12 \), and \( (3)(5) = 15 \).

Step 4: Sum the results to find the total work done

Combine the results from Step 3: \( W = -4 + 12 + 15 = 23 \).

Step 5: Determine the sign of the work done

Since the total is positive, the work done by the force is \(+23 \mathrm{~J}\). Therefore, the correct answer is \((a) +23 \mathrm{~J}\).

Key Concepts

Vector Displacement

When we talk about vector displacement, we are talking about a vector that represents the change in position of an object. This is different from just measuring the distance, as vector displacement considers both magnitude and direction. In the given exercise, we need to calculate the vector displacement from point \( P \) to point \( Q \). This is done using their coordinate points:

• Point \( P = (3, 3, -1) \)
• Point \( Q = (2, -1, 4) \)

We find the displacement vector \( \mathbf{d} \) by subtracting the coordinates of \( P \) from those of \( Q \), resulting in:
\( \mathbf{d} = (2 - 3) \hat{\mathbf{i}} + (-1 - 3) \hat{\mathbf{j}} + (4 - (-1)) \hat{\mathbf{k}} = -1 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} + 5 \hat{\mathbf{k}} \).
This concise expression captures not only how far the object moved, but in which direction it traveled along each axis.

Dot Product

The dot product, or scalar product, is a way to multiply two vectors to get a scalar (a real number). This operation is especially useful in physics to find work done, as shown in the exercise. To calculate the dot product of two vectors, you multiply their corresponding components and then sum these products.
In our exercise, we have a force vector \( \mathbf{F} = 4 \hat{\mathbf{i}} - 3 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}} \) and a displacement vector \( \mathbf{d} = -1 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} + 5 \hat{\mathbf{k}} \).
The dot product \( \mathbf{F} \cdot \mathbf{d} \) is computed as:

• \( (4)(-1) = -4 \)
• \( (-3)(-4) = 12 \)
• \( (3)(5) = 15 \)

Adding these products together gives us the work done: \(-4 + 12 + 15 = 23\).
The result, 23, is a scalar indicating the total work done by the force when moving the object between the points.

Force Vector

A force vector is a vector that depicts the magnitude and direction of a force acting on an object. In physics, understanding force vectors helps us analyze how objects interact under various forces. The force vector in the exercise is given by \( 4 \hat{\mathbf{i}} - 3 \hat{\mathbf{j}} + 3 \hat{\mathbf{k}} \) N, representing a three-dimensional force acting along the x, y, and z axes.
This means:

• There is a force component of 4 N in the positive x-direction.
• A force component of -3 N in the positive y-direction, which implies a force acting in the negative y-direction.
• A force component of 3 N in the positive z-direction.

By considering all three components together, the vector accurately describes how the force is applied in space, allowing us to calculate how much work it performs as it moves an object along a specific path.

Coordinate Points

Coordinate points are the foundation for describing positions in a multi-dimensional space. In this exercise, we use coordinate points to define locations of points \( P \) and \( Q \) in a 3D space.

• Point \( P = (3, 3, -1) \)
• Point \( Q = (2, -1, 4) \)

Each coordinate point consists of values along the x, y, and z axes. For instance, \( P = (3, 3, -1) \) implies 3 units along the x-axis, 3 units along the y-axis, and -1 unit along the z-axis. Coordinate points help us in determining the vector displacement by subtracting the initial point \( P \) from the final point \( Q \).
Understanding the transition between these points through coordinates is crucial for calculating other vector-derived quantities, like displacement and thus the work done, as shown in this problem.