Key Concepts and Formulas

• The work done ($W$) by a constant force $\mathbf{F}$ on a particle undergoing a displacement $\mathbf{d}$ is given by the dot product: $W = \mathbf{F} \cdot \mathbf{d}$.
• If $\mathbf{F} = F_x\widehat i + F_y\widehat j + F_z\widehat k$ and $\mathbf{d} = d_x\widehat i + d_y\widehat j + d_z\widehat k$, then $\mathbf{F} \cdot \mathbf{d} = F_xd_x + F_yd_y + F_zd_z$.
• When multiple constant forces act on a particle, the total work done is the sum of the work done by each force, or equivalently, the work done by the resultant force. The resultant force is the vector sum of individual forces: $\mathbf{F_{net}} = \mathbf{F_1} + \mathbf{F_2} + \dots + \mathbf{F_n}$.

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Step-by-Step Solution

Step 1: Determine the displacement vector of the particle. The displacement vector $\mathbf{d}$ is the difference between the final position vector $\mathbf{r_B}$ and the initial position vector $\mathbf{r_A}$.

• Why: Work is defined in terms of the change in position, which is the displacement.
• Calculation: $\mathbf{r_A} = \widehat i + 2\widehat j + 3\widehat k$ $\mathbf{r_B} = 5\widehat i + 4\widehat j + \widehat k$ $\mathbf{d} = \mathbf{r_B} - \mathbf{r_A} = (5\widehat i + 4\widehat j + \widehat k) - (\widehat i + 2\widehat j + 3\widehat k)$ $\mathbf{d} = (5-1)\widehat i + (4-2)\widehat j + (1-3)\widehat k$ $\mathbf{d} = 4\widehat i + 2\widehat j - 2\widehat k$

Step 2: Calculate the resultant force acting on the particle. The resultant force $\mathbf{F_{net}}$ is the vector sum of the individual forces $\mathbf{F_1}$ and $\mathbf{F_2}$.

• Why: The total work done by multiple forces is equal to the work done by their resultant force. This simplifies the calculation.
• Calculation: $\mathbf{F_1} = 4\widehat i + \widehat j - 3\widehat k$ $\mathbf{F_2} = 3\widehat i + \widehat j - \widehat k$ $\mathbf{F_{net}} = \mathbf{F_1} + \mathbf{F_2} = (4\widehat i + \widehat j - 3\widehat k) + (3\widehat i + \widehat j - \widehat k)$ $\mathbf{F_{net}} = (4+3)\widehat i + (1+1)\widehat j + (-3-1)\widehat k$ $\mathbf{F_{net}} = 7\widehat i + 2\widehat j - 4\widehat k$

Step 3: Calculate the total work done by the forces. The total work done $W_{total}$ is the dot product of the resultant force $\mathbf{F_{net}}$ and the displacement vector $\mathbf{d}$.

• Why: This is the definition of work done by a constant force.
• Calculation: $W_{total} = \mathbf{F_{net}} \cdot \mathbf{d}$ $W_{total} = (7\widehat i + 2\widehat j - 4\widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k)$ $W_{total} = (7)(4) + (2)(2) + (-4)(-2)$ $W_{total} = 28 + 4 + 8$ $W_{total} = 40$

Alternative Method (Sum of Individual Works):

Step 2 (Alternative): Calculate the work done by each force individually.

Step 2a: Work done by $\mathbf{F_1}$.

• Why: To find the contribution of each force to the total work.
• Calculation: $W_1 = \mathbf{F_1} \cdot \mathbf{d} = (4\widehat i + \widehat j - 3\widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k)$ $W_1 = (4)(4) + (1)(2) + (-3)(-2) = 16 + 2 + 6 = 24$

Step 2b: Work done by $\mathbf{F_2}$.

• Why: To find the contribution of each force to the total work.
• Calculation: $W_2 = \mathbf{F_2} \cdot \mathbf{d} = (3\widehat i + \widehat j - \widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k)$ $W_2 = (3)(4) + (1)(2) + (-1)(-2) = 12 + 2 + 2 = 16$

Step 3 (Alternative): Sum the individual works to find the total work.

• Why: The total work is the sum of work done by all forces.
• Calculation: $W_{total} = W_1 + W_2 = 24 + 16 = 40$

Both methods yield the same result.

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Common Mistakes & Tips

• Displacement Calculation: Ensure you subtract the initial position vector from the final position vector correctly to find the displacement. Mistakes in signs are common.
• Dot Product Calculation: Remember that the dot product involves summing the products of corresponding components. Forgetting the negative sign in a component product can lead to an incorrect answer.
• Vector Addition/Subtraction: Pay close attention to the signs when adding or subtracting vector components.

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Summary

The problem involves calculating the total work done by two constant forces acting on a particle that undergoes a displacement. We first determined the displacement vector by subtracting the initial position vector from the final position vector. Then, we found the resultant force by adding the two force vectors. Finally, the total work done was calculated as the dot product of the resultant force and the displacement vector. Alternatively, we could have calculated the work done by each force individually and summed them to obtain the total work. Both methods confirmed the total work done to be 40 units.

The final answer is $\boxed{40}$ units, which corresponds to option (D).