Question

A body of mass 2 kg is moving with a constant acceleration of \( \left( 2\hat{i} + 3\hat{j} - \hat{k} \right) \, \text{ms}^{-2} \). If the displacement made by the body is \( \left( 3\hat{i} - \hat{j} + 2\hat{k} \right) \, \text{m} \), then the work done is:

• A. 22 J
• B. 2 J
• C. 12 J
• D. 10 J

Hint:

To calculate work done, remember the formula \( W = \vec{F} \cdot \vec{d} \), and always find the dot product of the force and displacement vectors.

Answer 1

The Correct Option is B

The work done is given by the equation: \[ W = \vec{F} \cdot \vec{d} \] where \( \vec{F} \) is the force and \( \vec{d} \) is the displacement. The force is calculated from Newton’s second law: \[ \vec{F} = m \cdot \vec{a} \] where \( m = 2 \, \text{kg} \) and \( \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} \, \text{ms}^{-2} \). Thus, \[ \vec{F} = 2 \times (2\hat{i} + 3\hat{j} - \hat{k}) = 4\hat{i} + 6\hat{j} - 2\hat{k} \, \text{N} \] The displacement vector is given as: \[ \vec{d} = 3\hat{i} - \hat{j} + 2\hat{k} \, \text{m} \] Now, compute the dot product \( \vec{F} \cdot \vec{d} \): \[ \vec{F} \cdot \vec{d} = (4\hat{i} + 6\hat{j} - 2\hat{k}) \cdot (3\hat{i} - \hat{j} + 2\hat{k}) \] \[ = (4 \times 3) + (6 \times -1) + (-2 \times 2) \] \[ = 12 - 6 - 4 = 2 \, \text{J} \] Thus, the work done is \( 2 \, \text{J} \).

Answer 2

Step 1: Write down the given data
Mass of the body, \( m = 2 \, \text{kg} \)
Acceleration, \( \vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} \, \text{ms}^{-2} \)
Displacement, \( \vec{s} = 3\hat{i} - \hat{j} + 2\hat{k} \, \text{m} \)

Step 2: Find the force acting on the body
Using Newton's second law:
\[ \vec{F} = m \vec{a} = 2(2\hat{i} + 3\hat{j} - \hat{k}) = 4\hat{i} + 6\hat{j} - 2\hat{k} \, \text{N} \]

Step 3: Calculate the work done
Work done, \( W = \vec{F} \cdot \vec{s} \)
\[ W = (4\hat{i} + 6\hat{j} - 2\hat{k}) \cdot (3\hat{i} - \hat{j} + 2\hat{k}) = (4 \times 3) + (6 \times -1) + (-2 \times 2) = 12 - 6 - 4 = 2 \, \text{J} \]

Final answer: The work done by the body is 2 Joules.