JEE Main 2017 · 2 Apr · Shift 1 · Physics Q4

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A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be :

4.5 J

22 J

9 J

18 J

Answer and worked solution

Official answer: A

$a=6t$ , so $v=3t^2$ and $v(1)=3$ m/s. By the work-energy theorem, $W=\frac12(1)(3)^2=4.5$ J.

Topics:Work by constant and variable forces

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JEE Main 2017 · 2 Apr · Shift 1, Physics question 4