A uniform force of $(3\hat{i} + \hat{j})$ N acts on a particle of mass 2 kg. The particle is displaced from th

Question

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

Accepted Answer

According to the definition of work done by a constant force, work ($W$) is the scalar product of the force vector ($\mathbf{F}$) and the displacement vector ($\mathbf{d}$), expressed as $W = \mathbf{F} \cdot \mathbf{d}$ .

1. **Find the displacement vector ($\mathbf{d}$)**: Displacement is the difference between the final position ($\mathbf{r_2}$) and the initial position ($\mathbf{r_1}$) .
   $$\mathbf{d} = \mathbf{r_2} - \mathbf{r_1} = (4\hat{i} + 3\hat{j} - \hat{k}) - (2\hat{i} + \hat{k})$$
   $$\mathbf{d} = (4-2)\hat{i} + (3-0)\hat{j} + (-1-1)\hat{k} = 2\hat{i} + 3\hat{j} - 2\hat{k} 	ext{ m}$$

2. **Calculate the scalar product**: Multiply the corresponding components of the force $\mathbf{F} = (3\hat{i} + \hat{j} + 0\hat{k})$ N and the displacement vector .
   $$W = (3)(2) + (1)(3) + (0)(-2)$$
   $$W = 6 + 3 + 0 = 9 	ext{ J}$$

Thus, the work done by the force is **9 J**.

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