JEE Main · PhysicsMedium

$A force F = (3 x^{2} + 2 x) N acts on a particle moving along the x -axis. What is the work done by this force as the particle moves from x = 0 to x = 2 m ?$

A. $8 J$
B. $12 J$
C. $16 J$
D. $10 J$

Show correct answer & step-by-step solution

Correct answer: B — $12 J$

Solution

$Work done is W = \int_{0}^{2} (3 x^{2} + 2 x) d x .$
$Integrating the terms gives [x^{3} + x^{2}]_{0}^{2} .$
$Evaluating at the limits gives (2^{3} + 2^{2}) - (0^{3} + 0^{2}) .$
$W = 8 + 4 = 12 J .$
$Hence the answer is (B).$

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