Solution:
The work done by the force is \( W = \int_{2}^{3} 3x \, dx + \int_{3}^{0} 4 \, dy = \left[ \frac{3x^2}{2} \right]_2^3 + [4y]_3^0 = 7.5 - 12 = -4.5\text{ J} \).
A force \( \vec{F} = (3x\hat{i} + 4\hat{j})\text{ Newton} \) (where \( x \) is in metres) acts on a particle which moves from a position \( (2\text{ m}, 3\text{ m}) \) to \( (3\text{ m}, 0\text{ m}) \). Then the work done is:
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