Question

Work done by a moving particle in the force field \( \mathbf{F} = 6x^2 \hat{i} + (3xz + y) \hat{j} + 4z \hat{k} \), moving along the straight line from (0,0,0) to (1,2,3) is _____. \(\textit{[Answer in integer]}\)

Hint:

The work done in a force field can be calculated using the dot product of the force field and the displacement vector, and then integrating along the path of motion.

Answer 1

Correct Answer: 28

The work done by a moving particle is given by the formula: \[ W = \int_{\mathbf{r_1}}^{\mathbf{r_2}} \mathbf{F} \cdot d\mathbf{r} \] where: - \( \mathbf{F} = 6x^2 \hat{i} + (3xz + y) \hat{j} + 4z \hat{k} \) is the force field, - \( \mathbf{r_1} = (0,0,0) \) and \( \mathbf{r_2} = (1,2,3) \) are the initial and final positions, - \( d\mathbf{r} = dx \hat{i} + dy \hat{j} + dz \hat{k} \) is the infinitesimal displacement vector. The work integral will be evaluated along the straight line from (0,0,0) to (1,2,3). To parameterize the path, we write: \[ x = t, \quad y = 2t, \quad z = 3t \quad \text{where} \quad t \in [0,1]. \] The differentials are: \[ dx = dt, \quad dy = 2dt, \quad dz = 3dt. \] Substitute into the force equation: \[ \mathbf{F} = 6x^2 \hat{i} + (3xz + y) \hat{j} + 4z \hat{k}. \] Now, calculate the dot product \( \mathbf{F} \cdot d\mathbf{r} \): \[ \mathbf{F} \cdot d\mathbf{r} = (6x^2 \, dx) + (3xz + y) \, dy + (4z \, dz). \] Substitute the expressions for \( x \), \( y \), and \( z \) into the equation. After integrating over the given bounds from 0 to 1, we find that: \[ W = 28. \] Thus, the work done by the particle is \( \boxed{28} \) (in integer).