A paticle acted on by constant forces `4hati+hatj-3hatk and 3hati+hatj-hatk` is displaced from the point `hati+2hatj+3hatk to 5hati+4hatj+hatk`. Find the work done

Options

A. 50 units
B. 20 units
C. 30 units
D. 40 units

Explanation

To solve the problem step by step, we will follow the outlined process in the video transcript.

### Step 1: Identify the Forces
We have two constant forces acting on the particle:
- \( \vec{F_1} = 4\hat{i} + \hat{j} - 3\hat{k} \)
- \( \vec{F_2} = 3\hat{i} + \hat{j} - \hat{k} \)

### Step 2: Calculate the Net Force
The net force \( \vec{F} \) acting on the particle is the sum of the two forces:
\[ \vec{F} = \vec{F_1} + \vec{F_2} \]
Calculating this:
\[ \vec{F} = (4\hat{i} + \hat{j} - 3\hat{k}) + (3\hat{i} + \hat{j} - \hat{k}) = (4 + 3)\hat{i} + (1 + 1)\hat{j} + (-3 - 1)\hat{k} \]
\[ \vec{F} = 7\hat{i} + 2\hat{j} - 4\hat{k} \]

### Step 3: Identify the Initial and Final Positions
The initial position \( \vec{d_1} \) and final position \( \vec{d_2} \) of the particle are given as:
- \( \vec{d_1} = \hat{i} + 2\hat{j} + 3\hat{k} \)
- \( \vec{d_2} = 5\hat{i} + 4\hat{j} + \hat{k} \)

### Step 4: Calculate the Displacement
The displacement \( \vec{d} \) is given by the difference between the final and initial positions:
\[ \vec{d} = \vec{d_2} - \vec{d_1} \]
Calculating this:
\[ \vec{d} = (5\hat{i} + 4\hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = (5 - 1)\hat{i} + (4 - 2)\hat{j} + (1 - 3)\hat{k} \]
\[ \vec{d} = 4\hat{i} + 2\hat{j} - 2\hat{k} \]

### Step 5: Calculate the Work Done
The work done \( W \) is given by the dot product of the net force \( \vec{F} \) and the displacement \( \vec{d} \):
\[ W = \vec{F} \cdot \vec{d} \]
Calculating this:
\[ W = (7\hat{i} + 2\hat{j} - 4\hat{k}) \cdot (4\hat{i} + 2\hat{j} - 2\hat{k}) \]
Using the dot product:
\[ W = (7 \cdot 4) + (2 \cdot 2) + (-4 \cdot -2) \]
\[ W = 28 + 4 + 8 = 40 \]

### Final Answer
The work done is \( 40 \) units.

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