A uniform force of $(3\hat{i} + \hat{j}) ext{ N}$ acts on a particle of mass $2 ext{ kg}$. Hence the parti

Question

A uniform force of $(3\hat{i} + \hat{j}) 	ext{ N}$ acts on a particle of mass $2 	ext{ kg}$. Hence the particle is displaced from position $(2\hat{i} + \hat{k}) 	ext{ m}$ to position $(4\hat{i} + 3\hat{j} - \hat{k}) 	ext{ m}$. The work done by the force on the particle is:

Accepted Answer

1. **Definition of Work:** Work done ($W$) by a constant force ($\vec{F}$) displacing a particle is the dot product of the force vector and the displacement vector ($\vec{d}$).
   $$ W = \vec{F} \cdot \vec{d} $$
   (Refer to NCERT Class 11, Chapter 6, Section 6.3, Eq. 5.4).
2. **Calculate Displacement Vector ($\vec{d}$):** Displacement is the change in position vector, $\vec{d} = \vec{r}_f - \vec{r}_i$.
   - Initial Position ($\vec{r}_i$): $2\hat{i} + 0\hat{j} + \hat{k}$
   - Final Position ($\vec{r}_f$): $4\hat{i} + 3\hat{j} - \hat{k}$
   $$ \vec{d} = (4\hat{i} + 3\hat{j} - \hat{k}) - (2\hat{i} + 0\hat{j} + \hat{k}) $$
   $$ \vec{d} = (4-2)\hat{i} + (3-0)\hat{j} + (-1-1)\hat{k} $$
   $$ \vec{d} = 2\hat{i} + 3\hat{j} - 2\hat{k} 	ext{ m} $$
3. **Calculate Work Done:** Given Force $\vec{F} = 3\hat{i} + \hat{j} = 3\hat{i} + 1\hat{j} + 0\hat{k} 	ext{ N}$.
   $$ W = \vec{F} \cdot \vec{d} = (3)(2) + (1)(3) + (0)(-2) $$
   $$ W = 6 + 3 + 0 = 9 	ext{ J} $$
   (Refer to NCERT Class 11, Section 6.1.1 for Scalar Product rules).
4. **Note on Mass:** The mass of the particle ($2 	ext{ kg}$) is extraneous information for calculating work done in this context, as work depends only on force and displacement.

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