A body of mass 0.5 kg travels on straight line path with velocity v = (3x² + 4) m/s. The net workdone by the force during its displacement from x = 0 to x = 2 m is: (a) 64 J (b) 60 J (c) 120 J (d) 128 J

Answer: (b). Net work = change in KE. v(0) = 4 m/s, v(2) = 3(4) + 4 = 16 m/s W = 1/2(0.5)(16² - 4²) = 0.25 × 240 = 60 J So the answer is (b) 60 J.

A body of mass 0.5 kg travels on straight line path with velocity v =… | JEE Main 2022 (25 Jul, Shift 1)

Q12JEE Main 2022 · 25 JUL · SHIFT 1Medium

Work-Energy Theorem

A body of mass $0.5\text{ kg}$ travels on straight line path with velocity $v = \left(3x^2 + 4\right)\text{ m/s}$ . The net workdone by the force during its displacement from $x = 0$ to $x = 2\text{ m}$ is:

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Answer: (b)

Net work $=$ change in KE.
$v(0) = 4\text{ m/s}$ , $v(2) = 3(4) + 4 = 16\text{ m/s}$
$W = \frac{1}{2}(0.5)(16^2 - 4^2) = 0.25 \times 240 = 60\text{ J}$
So the answer is (b) 60 J.

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