A particle moves from a point $(-2\hat i+5\hat j)$ to $(4\hat j+3\hat k)$ when a force of $(4\hat i+3\hat j)$

Question

A particle moves from a point $(-2\hat i+5\hat j)$ to $(4\hat j+3\hat k)$ when a force of $(4\hat i+3\hat j)$ N is applied. How much work has been done by the force?

Accepted Answer

1. **Calculate Displacement Vector ($\mathbf{d}$):** The displacement is the difference between the final position vector ($\mathbf{r}_2$) and the initial position vector ($\mathbf{r}_1$).
   $$ \mathbf{r}_1 = -2\hat i + 5\hat j $$
   $$ \mathbf{r}_2 = 4\hat j + 3\hat k $$
   $$ \mathbf{d} = \mathbf{r}_2 - \mathbf{r}_1 = (0 - (-2))\hat i + (4 - 5)\hat j + (3 - 0)\hat k $$
   $$ \mathbf{d} = 2\hat i - 1\hat j + 3\hat k $$
2. **Identify Force Vector ($\mathbf{F}$):**
   $$ \mathbf{F} = 4\hat i + 3\hat j $$
3. **Calculate Work Done ($W$):** Work is defined as the scalar (dot) product of force and displacement.
   $$ W = \mathbf{F} \cdot \mathbf{d} $$ [Class 11 Physics, Ch 5, Sec 5.3, Eq 5.4]
   $$ W = (4\hat i + 3\hat j + 0\hat k) \cdot (2\hat i - 1\hat j + 3\hat k) $$
   Using the property $\hat i \cdot \hat i = 1, \hat i \cdot \hat j = 0$, etc.:
   $$ W = (4)(2) + (3)(-1) + (0)(3) $$
   $$ W = 8 - 3 + 0 = 5	ext{ J} $$ [Class 11 Physics, Ch 5, Sec 5.1.1]

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