Vector Algebra
Dot Product and Work
Grade None
Question:
<p>Two forces <span>\(\vec{f_1} = 3\vec{i} - 2\vec{j} + \vec{k}\)</span> and <span>\(\vec{f_2} = 2\vec{i} + 3\vec{j} - 5\vec{k}\)</span> acting on a particle at A move it to B. The work done, if the position vector of A and B are <span>\(-2\vec{i} + 5\vec{k}\)</span> and <span>\(3\vec{i} - 7\vec{j} + 2\vec{k}\)</span>, is</p>
<p>(a) 20 units</p>
<p>(b) 7 units</p>
<p>(c) 25 units</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Work done by a force is the dot product of the resultant force and the displacement vector. Find the resultant force first, then calculate the dot product with displacement.
Solution: Let $\vec{R}$ be the resultant of two forces $\vec{f_1}$ and $\vec{f_2}$ and $\vec{d}$ be the displacement. $\vec{R} = \vec{f_1} + \vec{f_2} = (3\hat{i} - 2\hat{j} + \hat{k}) + (2\hat{i} + 3\hat{j} - 5\hat{k}) = 4\hat{i} + \hat{j} - 4\hat{k}$ $\vec{d} = (3\hat{i} - 7\hat{j} + 2\hat{k}) - (-2\hat{i} + 5\hat{k}) = 5\hat{i} - 7\hat{j} - 3\hat{k}$ The total work done $= \vec{R} \cdot \vec{d}$ $= (4\hat{i} + \hat{j} - 4\hat{k}) \cdot (5\hat{i} - 7\hat{j} - 3\hat{k})$ $= 20 - 7 + 12 = 25 \text{ units}$ ∴ Answer is (c) 25 units.
Correct Answer: C