A body of mass 0.5 kg travels on straight line path with velocity v = (3x² + 4)m/s. The net work done by the force during its displacement from x = 0 to x = 2 m is :

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120 J

64 J

60 J

128 J

answer is B.

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Detailed Solution

$\begin{array}{l} v_{i} = 3 (0^{2}) + 4 = 4 ≅ x = 0 \\ v_{F} = 3 (2)^{2} + 4 ≅ x = 2 \\ = 16 \\ W = ΔK = \frac{1}{2} m (16^{2} - 4^{2}) \\ = \frac{1}{2} \times \frac{1}{2} (256 - 16) \\ = \frac{240}{4} = 60 J \end{array}$

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