A variable force, given by the 2-dimensional vector $\overset{⇀}{F} = (3 x^{2} \hat{i} + 4 \hat{j})$ acts on a particle. The force is in newton and x is in metre. What is the change in the kinetic energy of the particle as it moves from the point with coordinates (2, 3) to (3, 0) ? (The coordinates are in metres)

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–7 J

+7 J

+19 J

zero

answer is C.

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Detailed Solution

$W = \int \overset{⇀}{F} . \overset{⇀}{d s} = \int (3 x^{2} \hat{i} + 4 \hat{j}) . (d x \hat{i} + d y \hat{j})$
$W = \int_{2}^{3} 3 x^{2} d x + \int_{3}^{0} 4 d y = {[x^{3}]}_{2}^{3} + 4 {[y]}_{3}^{0}$

$W = (27 - 8) + 4 (0 - 3) = 19 - 12 = 7 J$

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