A particle acted upon by constant forces `4hati +hatj- 4 hatk` and `3hati + hatj - hatk` is displacment from the point `hati+ 2hatj+ hatk` to point `5hati + 4hatj +hatk`.Total work done by the forces in SI unit is :

Options

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

Explanation

To solve the problem, we need to find the total work done by the forces acting on a particle as it moves from one point to another. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Forces
The forces acting on the particle are:
- \( \mathbf{F_1} = 4\hat{i} + \hat{j} - 4\hat{k} \)
- \( \mathbf{F_2} = 3\hat{i} + \hat{j} - \hat{k} \)

### Step 2: Calculate the Net Force
To find the total force acting on the particle, we sum the two forces:
\[ \mathbf{F_{net}} = \mathbf{F_1} + \mathbf{F_2} = (4\hat{i} + \hat{j} - 4\hat{k}) + (3\hat{i} + \hat{j} - \hat{k}) \]
Calculating this gives:
\[ \mathbf{F_{net}} = (4 + 3)\hat{i} + (1 + 1)\hat{j} + (-4 - 1)\hat{k} = 7\hat{i} + 2\hat{j} - 5\hat{k} \]

### Step 3: Determine the Displacement Vector
The particle moves from the initial point \( \mathbf{S_1} = \hat{i} + 2\hat{j} + \hat{k} \) to the final point \( \mathbf{S_2} = 5\hat{i} + 4\hat{j} + \hat{k} \).
The displacement vector \( \Delta \mathbf{S} \) is given by:
\[ \Delta \mathbf{S} = \mathbf{S_2} - \mathbf{S_1} = (5\hat{i} + 4\hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) \]
Calculating this gives:
\[ \Delta \mathbf{S} = (5 - 1)\hat{i} + (4 - 2)\hat{j} + (1 - 1)\hat{k} = 4\hat{i} + 2\hat{j} + 0\hat{k} \]

### Step 4: Calculate the Work Done
The work done by the net force is given by the dot product of the net force and the displacement:
\[ W = \mathbf{F_{net}} \cdot \Delta \mathbf{S} \]
Substituting the values:
\[ W = (7\hat{i} + 2\hat{j} - 5\hat{k}) \cdot (4\hat{i} + 2\hat{j} + 0\hat{k}) \]
Calculating the dot product:
\[ W = (7 \times 4) + (2 \times 2) + (-5 \times 0) = 28 + 4 + 0 = 32 \text{ joules} \]

### Final Answer
The total work done by the forces is:
\[ \boxed{32 \text{ joules}} \]

---