A particle acted upon by constant forces `4hat(i) + hat(j) - 3hat(k) and 3 hat(i) + hat(j)- hat(k)` is displaced from the point `hat(i) + 2hat(j)- 3hat(k)` to point `5hat(i) + 4hat(j)- hat(k)`. The total work done by the forces in SI unit is

Options

A. 20J
B. 24J
C. 50J
D. 30J

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the Net Force
The net force \(\vec{F}_{net}\) acting on the particle is the vector sum of the two forces given:

\[ \vec{F}_1 = 4\hat{i} + \hat{j} - 3\hat{k} \]
\[ \vec{F}_2 = 3\hat{i} + \hat{j} - \hat{k} \]

Now, we add these two vectors:

\[ \vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = (4\hat{i} + \hat{j} - 3\hat{k}) + (3\hat{i} + \hat{j} - \hat{k}) \]

Combining like terms:

\[ \vec{F}_{net} = (4 + 3)\hat{i} + (1 + 1)\hat{j} + (-3 - 1)\hat{k} = 7\hat{i} + 2\hat{j} - 4\hat{k} \]

### Step 2: Calculate the Displacement
The displacement \(\vec{s}\) is given by the difference between the final and initial positions:

\[ \text{Initial position} = \hat{i} + 2\hat{j} - 3\hat{k} \]
\[ \text{Final position} = 5\hat{i} + 4\hat{j} - \hat{k} \]

Now, we calculate the displacement:

\[ \vec{s} = \text{Final position} - \text{Initial position} = (5\hat{i} + 4\hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} - 3\hat{k}) \]

Subtracting the vectors:

\[ \vec{s} = (5 - 1)\hat{i} + (4 - 2)\hat{j} + (-1 + 3)\hat{k} = 4\hat{i} + 2\hat{j} + 2\hat{k} \]

### Step 3: Calculate the Work Done
The work done \(W\) by the forces is given by the dot product of the net force and the displacement:

\[ W = \vec{F}_{net} \cdot \vec{s} \]

Substituting the values we found:

\[ W = (7\hat{i} + 2\hat{j} - 4\hat{k}) \cdot (4\hat{i} + 2\hat{j} + 2\hat{k}) \]

Calculating the dot product:

\[ W = (7 \cdot 4) + (2 \cdot 2) + (-4 \cdot 2) \]
\[ W = 28 + 4 - 8 \]
\[ W = 24 \, \text{Joules} \]

### Final Answer
The total work done by the forces is **24 Joules**.

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