A force ` bar F = (3hat i-4 hat j +b hat k)N is acting on the particle which is moving from point A ( 0-1, 1) m to the point B(2, 2 3) m If net work done by the force on the particle is zero then value of b is

Options

A. -3
B. -2
C. 2
D. 3

Answer

To solve the problem step by step, we need to find the value of \( b \) such that the net work done by the force \( \mathbf{F} = (3 \hat{i} - 4 \hat{j} + b \hat{k}) \, \text{N} \) on a particle moving from point \( A(0, -1, 1) \) m to point \( B(2, 2, 3) \) m is zero.

### Step 1: Determine the displacement vector \( \mathbf{ds} \)

The displacement vector \( \mathbf{ds} \) can be calculated as:

\[ \mathbf{ds} = \mathbf{r_f} - \mathbf{r_i} = (2, 2, 3) - (0, -1, 1) \]

Calculating the components:

- \( x \): \( 2 - 0 = 2 \)
- \( y \): \( 2 - (-1) = 2 + 1 = 3 \)
- \( z \): \( 3 - 1 = 2 \)

Thus,

\[ \mathbf{ds} = 2 \hat{i} + 3 \hat{j} + 2 \hat{k} \]

### Step 2: Calculate the work done \( W \)

The work done \( W \) by the force \( \mathbf{F} \) is given by the dot product of \( \mathbf{F} \) and \( \mathbf{ds} \):

\[ W = \mathbf{F} \cdot \mathbf{ds} = (3 \hat{i} - 4 \hat{j} + b \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} + 2 \hat{k}) \]

Calculating the dot product:

\[ W = (3)(2) + (-4)(3) + (b)(2) \]

This simplifies to:

\[ W = 6 - 12 + 2b = -6 + 2b \]

### Step 3: Set the work done to zero

Since we want the net work done to be zero:

\[ -6 + 2b = 0 \]

### Step 4: Solve for \( b \)

Rearranging the equation to solve for \( b \):

\[ 2b = 6 \]

Dividing both sides by 2:

\[ b = 3 \]

### Conclusion

The value of \( b \) for which the net work done by the force on the particle is zero is:

\[ \boxed{3} \]

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